Electrostatics

The branch of physics, deals with electric effect of static charge is called electrostatics.

1. Electric Charge
Charge of a body or particle is the property due to which it produces and experiences
electrical and magnetic effects.
–3 –6 –9
It is measured in coulomb in (S.Ι. unit). In practice we use mC (10 C), µC (10 C), nC(10 C) etc.
C.G.S. unit of charge = stat coulomb
= electrostatic unit = esu = franklin
9
1 coulomb = 3 × 10 esu of charge
0 0 1 1
Dimensional formula of charge = [M L T A ]

1.1 Properties of Charge

• Charge is a scalar quantity.
• Charge is of two types: (i) Positive charge and (ii) Negative charge.
Charging a body implies transfer of charge (electrons) from one body to another. Positively
charged body means loss of electrons, i.e., deficiency of electrons. Negatively charged body
means excess of electrons. This also shows that mass of a negatively charged body > mass of
a positively charged [for identical bodies] body.
• Charge is conserved: In an isolated system, total charge remains constant whatever change
takes place in that system.
• Charge is quantized: Charge on a body always exists in integral multiples of a fundamental
unit of electric charge. So, charge on a body is Q = ± ne, where n is an integer and e is the
charge of the electron.
• Like point charges repel each other while unlike point charges attract each other.
• Charge is always associated with mass, i.e., charge can not exist without mass though mass
can exist without charge.
• Charge is independent of frame of reference, i.e., the charge is invariant. Charge on a body
does not change whatever be its speed. The mass of a body depends on its speed and
increases with increase in speed.
m0
m= m0 = Rest mass and c = speed of light
1 – V2 / C2
• A charge at rest produces only electric field around itself; a charge having uniform motion
produces electric as well as magnetic field around itself while a charge having accelerated
motion emits electromagnetic radiation.

Example 1:
–7
When a piece of polythene is rubbed with wool, a charge of –2 × 10 C is developed on
polythene. What is the amount of mass, which is transferred to polythene.
Solution:
Q 2 × 10–7
n
no. of electrons = = =
e 1.6 × 10–19
12
= 1.25 × 10
Now mass of transferred electrons
= n × mass of one electron
12 –31 –19
= 1.25 × 10 × 9.1 × 10 ≈ 11 × 10 kg
Example 2:
12
The rate of alpha particle falls on neutral sphere is 10 per second. What is the time in which
sphere gets charged by 2µC ?
Solution:
12
Number of α – particles falls in t second = 10 t
Charge on α – particle = +2e,
12
So charge incident in time t = (10 t).(2e)
Given charge is 2 µC
–6 12
∴ 2 × 10 = (10 t)·(2e)
10–18
=
⇒ t = 6.25 sec.
1.6 × 10–19

1.2 Charging of a body

A body can be charged by means of (A) friction, (B) conduction and (C) induction
(A) Friction:
When we rub a body with the other, the heat energy removes the loosely bounded electrons
from one body and the liberated electrons may enter into the other body. In consequence,
one body becomes positively charged by losing electrons and the other will get a net negative
charge by gaining electrons. for e.g., (a) when a glass rod is rubbed with silk, the glass gets a
positive charge and the silk becomes negatively charged. (b) Atmospheric electricity such as
charge deposition in clouds occur due to its friction with air.
(B) Conduction
When we touch a charged body (a charged conductor) with barefoot charge flows from the
body to earth through our body. This process is called conduction (flow) of charge. If we
touch (or connect) a neutral metallic ball to a charged metallic ball, the neutral ball will be
charged by sharing of charges between them by the process of conduction.
Note: If two conducting spheres are connected then the net charge on the spheres get divided
in the ratio of their radius. (Q ∝ R)
(C) Induction
When a neutral object B is kept near a positively charged object A, the neutral object induces
a negative charge nearer to the charged object and equal positive charge at its opposite side.
If the object B is a conductor, by connecting it with earth, excess electrons flow from earth to
neutralise the induced positive charges leaving it as a negatively charged object. This process
of charging a neutral object without touching it to a charged object is called “Induction”.
If there is attraction between two objects then one of them may be neutral. But if there is
repulsion between two objects, both must be charged (similarly charged).
So "repulsion is the sure test of electrification".

Example 3:
If a charged body is placed near a neutral conductor, will it attract the conductor or repel it?
Solution:
If a positive charged body is placed left-side near a neutral
conductor, negative charge will induce at left surface and –– +
––
–
+
positive charge will induce at right surface. Due to positively + q
– +
– + Repulsion
charged body negative induced charge will feel attraction and – +
Attraction – force
the positive induced charge will feel repulsion. But as the force – +
+
negative induced charge is nearer, so the attractive force will
be greater than the repulsive force. So the net force on the
conductor due to positively charged body will be attractive. Similarly, we can prove for
negatively charged body also.
 Detection of Charge : Electroscope
Let us investigate a way of determining the sign of the charge on an
object. An electroscope is one of the simplest device used to
Metal Knob
determine electric charge. An uncharged electroscope can detect
only whether an object is electrically charged.
Metal Rod
If the electroscope is previously charged with a known sign, it can
then also determine the sign of the charge on the object. An Strips
electroscope consists of two strips of thin aluminum or gold foil
fastened to a metal rod and connected to the knob at the top. When
a body touches the metal knob at the top of the electroscope, the
leaves diverge, indicating that the body is charged. If they do not
diverge then body is uncharged.

Example 4:
A glass rod rubbed with silk is used to charge a gold leaf electroscope and the leaves are
observed to diverge. The electroscope thus charged is exposed to X-ray for a short period.
Then
(1) The divergence of leaves will not be affected
(2) The leaves will diverge further
(3) The leaves will collapse
(4) The leaves will melt
Ans. (2)
Solution:
Charge on glass rod is positive, so charge on gold leaves will also be positive. Due to X-rays,
more electrons from leaves will be emitted, so leaves become more positive and diverge
further.

2. Coulomb's law (Inverse Square Law)

The magnitude of electrostatic force between two-point charges is directly proportional to
the product of charges and inversely proportional to the square of the distance between
them.
1 qq Kq1q2
i.e. F ∝ q1q2 and F ∝ 2 ⇒ F ∝ 12 2 ⇒F =
r r r2
Important points regarding Coulomb's law:
• It is applicable only for point charges. It is because the distribution of charge does not remain
uniform when two bodies are brought together.
1
• The constant of proportionality K in SI units in vacuum is expressed as and in any other
4πε0
1
medium expressed as [ε =ε0 .εr ] .
4πεM M
1 q1q2
• If charges are dipped in a medium then electrostatic force on one charge is . ε0 and
4πε0 εr r2
εM are called permittivity of vacuum and absolute permittivity of the medium respectively. The
ratio εM / ε0 =εr is called relative permittivity of the medium, which is a dimensionless
quantity.
• The value of relative permittivity εr is constant for medium and can have values between 1 to
∞. For vacuum, by definition it is equal to 1. For air it is nearly equal to 1 and may be taken to
be equal to 1 for calculations. For metals the value of εr is ∞.
 1 9 2 –2 –12 2 2
• The value of = 9 × 10 Nm C ⇒ ε0 = 8.855 × 10 C /Nm
4πε0
–1 –3 4 2
Dimensional formula of permittivity is M L T A
• The force acting on one point charge due to the other point charge is always along the line
joining these two charges. It is equal in magnitude and opposite in direction on two charges,
irrespective of the medium, in which they lie.
• The force is conservative in nature i.e., work done by electrostatic force in moving a point
charge along a closed loop of any shape is zero.
 1 q1q2 1 q1q2 
• vector form = F =  ˆr  r
4πε0 εr | r | 4πε0 εr | r |3

(q1 & q2 are to be substituted with sign) here r is position vector of the test charge (on which
force is to be calculated) with respect to the source charge (due to which force is to be
calculated).
• The force exerted by one charge on another is independent of medium & presence of other
charge.

Example 5:
Two particles having charges q1 and q2 when kept at a certain distance, exert a force F on
each other. If the distance between the two particles is reduced to half and the charge on
each particle is doubled then what will be the force between the particles:
Solution:
kq1q2 r
 F= If q’1 = 2q1, q’2 = 2q2, r’ =
r 2
2
kq'1 q'2 k(2q1 )(2q2 ) 16kq1q2
then F’ = = F’ = ⇒ F’ = 16F
r '2 r
2
r2
 
2

Example 6:
A particle of mass m carrying charge q1 is revolving around a fixed charge –q2 in a circular
path of radius r. Calculate the period of revolution and its speed also.
Solution:
1 q1q2 4π2mr
= mrω2 =
4πε0 r 2
T2

(4πε0 )r2 (4π2mr) πε0mr

T2 = or T = 4πr
q1q2 q1q2
and also we can say that
q1q2 mv 2 q1q2
= ⇒v=
4πε0r 2
r 4πε0mr

Example 7:
Two copper balls, each weighing 10g are kept in air 10 cm apart. If one electron from every
6
10 atoms is transferred from one ball to the other, the coulomb force between them is
(atomic weight of copper is 63.5)
10 4 8 6
(1) 2.0 × 10 N (2) 2.0 × 10 N (3) 2.0 × 10 N (4) 2.0 × 10 N
Solution:
e–
Number of atoms in given mass
+ –
10 23 22
= × 6.02 × 10 = 9.48 × 10
63.5 A B
10 cm
Number of electrons transferred between balls
9.48 × 1022 16
= = 9.48 × 10
106
Hence magnitude of charge gained by each ball
16 –19
Q = ne = 9.48 × 10 × 1.6 × 10 = 0.015 C
Force of attraction between the balls
9 (0.015)2 8
F = 9 × 10 × = 2 × 10 N
(0.1)2

Concept Builder-1
–20
Q.1 When a piece of a material is rubbed with another material, mass of 9.1 × 10 kg is reduced
from one material. Calculate the number of electrons gained by another material.
10 13 5 11
(1) 10 (2) 10 (3) 10 (4) 10

23
Q.2 Estimate the negative charge possessed by 1g of water. Given that Avogadro number 6.02×10
and molecular weight of water = 18.
4 4 4 4
(1) 6.25 × 10 C (2) 5.35 × 10 C (3) 1.76 × 10 C (4) 1.25 × 10 C

Q.3 Two point charges +3µC and 8µC are repel each other by force of 40 N. If a charge of –5µC is
added to each of them and the distance become half then what will be the force between
them?

Q.4 Two point charges +6µC and –12µC attract by the forces of 100 N. If the charges are increased
by 50% and the distance is reduced by 20% then what will be the new force between them ?

125
Q.5 A charge (q1 = 32µC) is placed at origin & another charge of magnitude µC & opposite
9
nature is placed at point (3, 4). The force acting on the second charge will be:

(
(1) 32 × 10–3 3iˆ + 4ˆj N ) ( )
(2) –32 × 10–3 3ˆi + 4ˆj N

(3) 32 × 10–3 (6iˆ + 8ˆj) N (

(4) –32 × 10–3 6ˆi + 8ˆj N )
Q.6 Find minimum repulsive force between 2 electrons placed at separation 1 fermi (1F = 10–15 m)
(1) 23 N (2) 2.3 N (3) 0.23 N (4) 230 N

3. Principle of Superposition 
Force on a charged particle due to number of point charges is the F4 
q1
resultant of forces due to individual point charges, therefore force on a F3

point test charge due to many charges is given by q2 F2
    q3 q
F = F1 + F2 + F3 + ....... . 
q4 F1
Example 8:
2 10
Two charged pendulums of charges µC & µC are in equilibrium when they are 1 m Apart.
3 3
If the pendulums are identical then what will be the mass of each pendulum if the angle
between the strings in equilibrium is 60°
Solution:
Force between the charges
2   10 
(9 × 109 )  × 10–6   × 10–3 
 3  3  T cos30º 60º
F=
12 30º
F = 20 N F = 20N
For balancing
T sin30º
⇒ T sin 30° = 20
⇒ T cos 30° = mg mg

20
⇒ tan 30° =
mg

⇒ m = 2 3 Kg

Concept Builder-2

Q.1 Three charges each of magnitude q are placed at the corners of an equilateral triangle, the
electrostatic force on the charge placed at the centre is (each side of triangle is L).
1 q2 1 3q2 1 q2
(1) zero (2) (3) (4)
4πε0 L2 4πε0 L2 12πε0 L2

Q.2 Charges Q, q, Q, q are placed at the corners A, B, C, D of a square respectively. If the resultant
force on the charge Q is zero due to other charges, what is the relation between Q and q ?
1
(1) Q = –2 2q (2) Q = –2q (3) Q = – 2q (4) Q = – q
2 2

Q.3 ABC is right-angle triangle with sides AB = 3 cm, BC = 4 cm, AC = 5 cm. A

+15
Charges 15, 12 and –20 e.s.u. are placed at A, B, C respectively.
5
Magnitude of the force experienced by the charge at B in dyne is: 3
(1) 125 (2) 35
B C
(3) 25 (4) zero +12 4 –20

Q.4 Two small balls having equal positive charge Q (coulomb) on each
are suspended by two insulated strings of equal length L meter, L
from a hook fixed to a stand. The whole set up is taken in satellite L
into space where there is no gravity (state of weightlessness). +Q +Q
Then the angle between the strings and tension in the string is
1 Q2 1 Q2
(1) 180°, . (2) 90°, . 2
4πε0 (2L)2 4πε0 L

1 Q2 1 QL2
(3) 180°, . 2 (4) 180°, . 2
4πε0 2L 4πε0 4L
Q.5 Two small spherical balls each carrying a charge Q = 10µC are suspended
by two insulating threads of equal length 1m each from ceiling.
60º
It is found that in equilibrium, threads are separated by an angle
60° between them, as shown in the figure.
1 9 2 Q Q
What is the tension in the threads (Given: = 9 × 10 Nm/C )
(4πε0 )
(1) 18 N (2) 1.8 N (3) 0.18 N (4) None

4. Electric Field
Electric field is the region around charged particle or charged body in which if another charge
is placed, it experiences electrostatic force.

4.1 Electric field intensity E
Electric field intensity at a point is equal to the electrostatic force experienced by a unit
positive point charge, both in magnitude and direction.

If a test charge q0 is placed at a point in an electric field and experiences a force F due to
some charges (called source charges), the electric field intensity at that point due to source

 F
charges is given by E = Lim
q →0 q
0 0


4.2 Properties of electric field intensity E
• It is a vector quantity. Its direction is the same as the force experienced by positive charge.
• Its S.Ι. unit is Newton/Coulomb [N/C].
–3 –1
• Its dimensional formula is [MLT A ].
• Electric force on a charge q placed in a region of electric field at a point where the electric
  
field intensity is E is given by F = qE .
• Electric force on point charge is in the same direction of electric field on positive charge and
in opposite direction on a negative charge.
• It obeys the superposition principle, that is, the field intensity at a point due to a system of
charges is vector sum of the field intensities due to individual point charges.
   
E = E1 + E2 + E3 + .....

5. Electric Lines of Force (ELOF) or Electric Field Lines (ELF)

An electric field line is an imaginary curve and tangent at any point on which gives the
direction of the net electric field at that point.
Electric lines of electrostatic field have following properties
(i) They are Imaginary
(ii) Can never intersect each other because there will be two directions of electric field at the
point of intersection which is not possible.
(iii) Can never forms closed loops because it originate from positive charge and terminate at
negative charge.
(iv) Electric field lines are always perpendicular to surface of metallic bodies.
(v) Lines of force starts from (+ve) charge or infinity and ends on (-ve)
charge or infinity.
(vi) If there is no electric field there will be no lines of force.
(vii) Lines of force per unit area normal to the area at a point represents
magnitude of intensity, crowded lines represent strong field while
distant represents weak field.
(viii) Tangent to the line of force at a point in an electric field gives the direction of intensity.
Example 9:
If number of electric lines of force from charge q are 10 then find out number of electric lines
of force from 2q charge.
Solution:
No. of ELOF ∝ magnitude of charge
10 ∝ q ⇒ 20 ∝ 2q So number of ELOF will be 20.

Example 10:
Some electric lines of force are shown in figure, for point A and B A
(1) EA > EB (2) EB > EA
B
(3) EA = EB (4) can’t be determined
Solution:
lines are denser at A so EA > EB.

Example 11:
If a charge is released in electric field, will it follow lines of force?
Solution:
Case I: F = q0E E
+q0
If lines of force are parallel (in uniform electric field):

In this type of field, if a charge is released, force on it will be qoE and its
→
direction will be along E . So the charge will move in a straight line along
the lines of force. +q
Case II: -
If lines of force are curved (in non-uniform electric field)
The charge will not follow lines of force

6. Electric Field Due to a Point Charge

kq
E= q r
r2
A P
 kq 
E= 3 r
r
  
• r = rP – rA = Position vector of P with respect to A. Put value of charge with sign.

Example 12:
Electrostatic force experienced by –3µC charge placed at point 'P' due
to a system 'S' of fixed point charges as shown in figure is P

= F (21ˆi + 9 ˆj) µN. Q1
Q2 S
(i) Find out electric field intensity at point P due to S.
(ii) If now 2µC charge is placed and –3µC is removed at point P then Q3 Q4
force experienced by it will be.

Solution:
   
(i) F = qE ⇒ (21iˆ + 9ˆj)µN = –3µC(E) ⇒ E = – 7 î – 3 ˆj
 
(ii) F2µC = +2( E ) = 2(–7 î – 3 ˆj ) = (–14 î – 6 ˆj ) µN
Example 13:
Calculate the electric field intensity which would be just sufficient to balance the weight of a
–2
particle of charge –10 µc and mass 10 mg. (take g = 10 ms )
Solution: Fe
 
| Fq | = | W |
i.e., |q|E = mg A q E
mg
⇒E = = 10 N/C in downward direction
|q|
W
Example 14:
Find out electric field intensity at point A (0, 1m, 2m) due to a point charge –20 µC situated at
point B( 2 m, 0, 1m).
Solution:
KQ 
E=  3r
|r|

⇒ r = Position vector of A – Position vector of B
   
r = rA – rB = (- 2 î + ˆj + k̂ ) ⇒ | r|= ( 2)2 + (1)2 + (1)2 = 2
9 × 109 × (–20 × 10–6 )
(– 2iˆ + ˆj + kˆ ) = – 22.5 × 10 (– 2iˆ + ˆj + kˆ ) N/C
3
E=
8

6.1 Null Point Identification

Case:1 For charges of same nature and unequal magnitude Null point will lie on the line of
joining of charges and in between the two charges
P
+ q1 r1 r2 + q2

at point ‘p’
kq1 kq2 r1 q
⇒ = ⇒ =1
r12 r22 r2 q2
Case:2 For charges of opposite nature & different magnitude Null point will lie on the line of
joining of the charges, closer to smaller magnitude charge & will be on exterior point
here |q1| > |q2|
–q2
P
q1 E=0
r1 r2

r1 q1
Here =
r2 q2

Example 15:
What has to be the value of Q for the system to be in equilibrium
q q
Q

q q
Solution:
Charge Q is at centre where E = 0
If system is in equilibrium then 'q' must be in equilibrium
Kq2 Kq2 2KQq FB F'centre
FD = FB = ; FA = , Fcentre =
a 2
2a 2
a 2 FA
D q
For equilibrium q FD
C
2KQq  Kq2 Kq2  Q
= = – 2 + 2 2 
a 2  2a a 
 q q
q  q q  A B
⇒ 2Q = –  + 2q  ; Q = –  + 
 2   4 2

Example 16: +q +q
Six equal point charges are placed at the corners of a regular
hexagon of side 'a'. Calculate electric field intensity at the centre of
hexagon? +q +q

Solution:
Zero +q +q

Example 17:
Calculate the electric field at origin due to infinite number of charges as shown in figures
below.
q q q q –q q
O O
1 2 4 x(m) 1 2 4 x(m)
fig(a) fig(b)
Solution:
1 1 1 
(a) E=
0
kq  + + + .....
 1 4 16 
kq.1 4kq
= =
(1 – 1 / 4) 3
a 1
[ S∞ = , a = 1 and r = ]
1–r 4
1 1 1  kq.1 4kq
E0 kq  – +
(b)= + ..... = =
 1 4 16  (1 – (–1 / 4)) 5

Concept Builder-3

Q.1 The lines of force of the electric field due to two charges q and Q are sketched in the figure
then:

Q q

(1) Q is positive and |Q| > |q| (2) Q is negative and |Q| > |q|
(3) q is positive and |Q| < |q| (4) q is negative and |Q| < |q|
Q.2 An uncharged sphere of metal is placed in between two charged plates as shown. The lines of
force look like
+ + + + + ++ – –– –––– + + + + + ++

– –– –––– + + + + + ++ – –– ––––
A B D
(1) A (2) B (3) C (4) D

Q.3 The given figure gives electric lines of force due to two charges q1 and q2. What are the signs
of the two charges ?

q1 q2

(1) Both are negative (2) Both are positive

(3) q1 is positive but q2 is negative (4) q1 is negative but q2 is positive
Q.4 Two charges +5µC and +10µC are placed 20 cm apart. The net electric field at the mid-Point
between the two charges is:
6 6
(1) 4.5 × 10 N/C directed towards + 5µC (2) 4.5 × 10 N/C directed towards +10µC
6 6
(3) 13.5 × 10 N/C directed towards +5µC (4) 13.5 × 10 N/C directed towards +10µC

Q.5 A charged particle of charge Q, mass m is kept in equilibrium in the electric field between the
plates of Millikan oil drop experiment. If the direction of the electric field between the plates
is reversed, then calculate acceleration of the charged particle.

Q.6 A charge q = 1 µC is placed at point (1m, 2m, 4m). Find the electric field at point P(0, –4m,
3m).

Q.7 A point charge of 0.009 µC is placed at origin. Calculate intensity of electric field due to this
point charge at point ( 2, 7,0) .

Q.8 The distance between the two charges 25µC and 36µC is 11cm. At what point on the line
joining the two, the intensity will be zero.
(1) At a distance of 5 cm from 25µC (2) At a distance of 5 cm from 36µC
(3) At a distance of 10 cm from 25µC (4) At a distance of 11 cm from 36µC

Q.9 Two point charges +8q and –2q are located at x=0 and x=L respectively. The location of a
point on the x-axis at which the net electric field due to these two point charges is zero is:
L
(1) 8L (2) 4L (3) 2L (4)
4
Q.10 For systems to be in equilibrium Q=?
q
Q
(i) a a (ii) q a a q
Q
q q
a
7. Electric Field Intensity Due To Continuous Charge Distribution
7.1 Electric field due to uniformly charged circular ring at an axial point
λRx kQx
=EP = 3/2 Q =λ(2πR)
2 ∈0 R + x 
2 2 r3
+ +
+ R + r 
2 2Q + + E
where; r= R +x ; λ = + O + x
P
2πR + +
kQx + +
or EP = +++
(R )
3/2
2
+ x2
At centre of ring x = 0 so E0 = 0
dE
E will be maximum when =0
dx
R 2KQ
that is at x = ⇒ Emax =
2 3 3 R2

KQ 2 2 2
Case (A): if x>>R, E = 2
, Hence the ring will act like a point charge (distant point x +R  x )
x
KQx
Case (B): if x<<R, E = 3
(nearpoint x2 + R2  R2 )
R
Example 18: +Q Q0
If Ep = 0, What is Q0? The distance of point P from C1 is same as C1 C2
P
that of C2.
R R
Solution: d

KQx K(Q0 )x
2 2 3/2
= ⇒ Q = Q0
(R + x ) (R2 + x2 )3/2

7.2 Segment of ring +λ

E at center +
+
+
2kλ α + 
E0 = sin α E
R 2 +
+
+ R
+
+
+λ

Example 19:
Electric field intensity at origin is – Y
λ 0– R
–
2Kλ0 2Kλ0 ˆ 45º
(1) î (2) – i – 45º
R R – X
R
2Kλ0 ˆ –
(3) – i (4) – x axis
R
Solution:
2Kλ0 2Kλ0
E= sin(45)(–i) ; E = (–i)
R R
7.3 Electric field due to uniformly charged infinite sheet
σ
Enet = toward normal direction
2ε0
• The direction of electric field is always perpendicular to the sheet.
• The magnitude of electric field is independent of distance from sheet.

8. Motion of Charge in A Uniform Electric Field

8.1 A charge is released from rest in a uniform electric field, then

• Force on charge = QE
QE
• Acceleration of charge =
m
E
Q⊕ F
QE
• Velocity after time ‘t’ = t
m
1 QE 2
• displacement after time ‘t’ = t
2 m
• Path of particle will be straight line

8.2 A charge enters in a uniform electric field with a velocity v parallel to the electric field

• Path of particle will be straight line

QE
• Acceleration of particle will be = Q⊕ v
m E
QE
• Velocity after time ‘t’ = v + t
m
1 QE 2
• Displacement after time ‘t’ = vt + t
2 m
• If the charge is enters along the field or positive charge enters opposite to the field then the
speed of particle will first decrease then increase.

v Q E

8.3 A charge enters in a uniform electric field perpendicularly

• Path of particle will be parabolic
• acceleration of particle
QE ˆ
a= j ⊕
v0
m
• velocity of particle after time ‘t’ E
ĵ
QE ˆ QE ˆ
= v 0ˆi + (t)
= j v 0ˆi + tj î
m m
9. Electric Flux (φ)
Electric flux is defined as ‘the total number of lines of force passing through normal area or
surface held perpendicularly’.

for open surface φ = ∫ dφ = ∫ E ds
for closed surface
 
φc =∫ E.ds (by definationof flux)
qnet
=φc or 4πkqnet (Gauss's law)
∈0

φc = E.S = E S cosθ

S = always normal to surface and pointed out wards.
(i) It is a scalar quantity
2 3 -3 -1
(ii) Units (V-m) and N-m /C, Dimensions: [ML T A ]
(iii) The value of φ does not depend upon the distribution of charges and the distance between
them inside the closed surface.
(iv) The value of φ is zero in the following circumstances:
(a) If a dipole (or many dipoles are) is enclosed by a closed surface
(b) Magnitude of (+ve) and (-ve) charges are equal inside a closed surface
(c) If no charge is enclosed by the closed surface
(d) Incoming flux (- ve) = out going flux (+ve)

9.1 Some Important application of flux linked with surface when electric field is uniform

ds 
2 2
(A) φin= –πR E and φout= πR E E

E
So φtotal = 0 R
 
ds  ds
E

2 2
(B) φin= φcircular = –πR E and φout= φcurved = πR E R 
E
So φtotal = 0

E


y 
(C) φin =−a2E and φout =
a2E ds E
So φtotal = 0  a
ds 
ds
x
a
a
z

1 1 
(D) φin= – πR2 E and φout = πR2E E
2 2
So φtotal = 0
R

Note: Electric flux linked with any closed surface in a uniform electric field is always zero.
10. Gauss's Law
1
The total flux linked with a closed surface is times the charge enclosed by the closed
∈0
  q
surface (Gaussian surface) i.e. ∫ E.ds = ∈ . This law is suitable to calculate for symmetrical
0
charge distribution and valid for all vector fields obeying inverse square law.
10.1 Gaussian surface
(i) Is taken to be imaginary closed surface
(ii) Is taken to be spherical for a point charge, conducting and non - conducting spheres
(iii) Is taken to be cylindrical for Infinite line of charges, uniformly charged long cylindrical bodies.

We select a Gaussian surface such that determination of

 
∫ E . ds can be done in simplest way (by symmetry)
Note: (i) Flux through Gaussian surface is independent of its shape.
(ii) Flux through Gaussian surface depends only on total charge present inside the Gaussian
surface.
(iii) Flux through Gaussian surface is independent of position of charges inside the Gaussian
surface.
(iv) Electric field intensity at the Gaussian surface is due to all the charges present inside as
well as outside the Gaussian surface.
(v) In a closed surface incoming flux is taken negative while outgoing flux is taken positive,
because n̂ is taken positive in outward direction.
(vi) In a Gaussian surface φ = 0 does not imply E = 0 at every point of the surface but E = 0 at
every point implies φ = 0.
(vii) Discrete charge can't present on Gaussian surface but continuous charge system can
pass through Gaussian surface.

10.2 Calculation using symmetry

(1) Charge placed at centre of sphere
(a) Flux linked with hemisphere 
kq   E
E = 2 ; φ = ∫ E· dA
R
q q
φ = 2πR2 × = R q
4π ∈0 R2 2 ∈0
Note: Here electric field is radial
(b) Flux linked with complete sphere

q
φTotal =
∈0 q

(2) Charge placed at centre of cylinder

(a) Flux linked with complete cylinder

q
φTotal =
∈0 q

(b) Flux linked with half of cylinder q

q
φcylinder =
2 ∈0
R
 (3) Charge placed in a cube
(i) Charge present at any one of the face of a
cube
q
q
φcube =
2 ∈0

(ii) Charged placed at centre

q
q
φTotal =
∈0

(iii) Charge placed at any one corner]

q
φ= q
8 ∈0
q
(iv) Charge is anyone arm of cube
q
φ=
4 ∈0

Example 20: closedsurface conductor

As shown in fig. a closed surface intersects a spherical conductor. If a
P
–ve charge is placed at point P. What is the nature of the electric flux
–Q
coming out of the closed surface ?
Solution: closedsurface conductor
Point charge –Q induces charge on conductor as shown in fig. Net – ++
charge enclosed by closed surface is (–ve) so flux is negative. –– + P
– +
– + –Q
–
– +
– +
– –+ +
Example 21:
Find out flux through the given Gaussian surface. • q5 =2µC
• q4 = –6µC • q1 =2µC
Solution:
• q2 =−3µC
Q 2µC – 3µC + 4µC 3 × 10–6 2
φ = in = = Nm /c • q3 =4µC
ε0 ε0 ε0 • q6 =3µC

Gaussian surface
11. Calculation of Electric field (Application of gauss law)
From gauss's law, we can say
 q
∫ E.ds = φnet = εin
0

11.1 E due to a uniformly charged spherical shell
(A) At Exterior Point
Since, electric field due to a shell will be radially outwards. So, let’s choose a spherical
Gaussian surface. Applying Gauss's law for this spherical Gaussian surface
 qin q q
∫ E ds = φnet = ε = ε ⇒ ∫ Eds cos 0° = q,R
o o
ε0 ++
+
+ +
2 qin q kq 1 + +
⇒ E (4πr ) = ⇒ Eout = = ∝ + r
ε0 4πεor2 r2 r2 + +
+ + +
 (B) At Interior Point
Let’s choose a spherical Gaussian surface inside the shell. Applying Gauss`s law for this
surface
 qin
∫ E.ds = φnet = ε = 0 ++
q,R
0 +
  + r +
Since angle between E and ds is 0°
  + +
Eds = E(4πr2 ) = 0 Ein = 0 + r

+ +
+
+ +


E

−R 
r
R

11.2 Electric field due to solid sphere (having uniformly distributed charge Q and radius R)
(A) At Exterior Point
Direction of electric field is radially outwards, so we will choose a spherical Gaussian surface
Applying Gauss's law
 qin Q
∫ E.ds = φnet = ε = ε
o o
  Q,R
Since angle between E and ds is 0° +
++ +
+ +
+ + +
  Q Q kQ 1 +
E.ds = E(4πr2 ) = ; Eout = = 2 ∝ 2 –

ε0 4πεor 2
r r

(B) At interior point

For this choose a spherical Gaussian surface inside the solid
sphere. Applying Gauss's law for this surface
++
+ + + Q,R +
r + ++
  + ++
+ + ++ + + +
 Q 4 3 ++
+ + +
+ +++
+
 × πr  + + + + + + ++
 4 πR3 3
+
 + + + + + ++ ++
 +++++ + ++ +
Qin 3
  = Qr
3
+ +
∫ E.ds = φnet =
ε0
=
ε0 εoR3

Qr kQr
E= ⇒ Ein = ∝r
4πεoR 3
R3

E

−R 
r
R
11.3 Electric field due to infinite line charge (having uniformly distributed charge of charge density λ)
Electric field due to infinite wire is radial so
we will choose cylindrical Gaussian surface as
shown is figure. ∞

φnet = φ1 + φ2 + φ3 φ1 = φ2 = 0(θ=90°) φ1
Surface(1)
cylindrical 
gaussian E
surface
qin λ
φnet = φ3 = = λ
εo εo  φ3
r
Surface(3) 
 r
R
and φ3 = ∫ E.ds
= ∫ E ds cos =
θ E∫ ds
= E(2πr )
φ2 Surface(2)

λ λ 2kλ 1 ∞
so, E (2πrλ) = ⇒E= = ∝
εo 2πεor r r

11.4 Electric field due to infinity long charged tube (having uniform surface charge density σ and
radius R)
 ∞
(A) E out side the tube
Let’s choose a cylindrical Gaussian surface. Applying Gauss’s r
theorem
+ + +
+ + +
qin σ2πR σ2πR + + +
φnet = = ⇒ Eout × 2πrλ = + + + 
εo εo εo + + +
+ + +
+ + +
σR 1
⇒E= ∝
r εo r

(B) E inside the tube: E
∞
Let’s choose a cylindrical Gaussian surface inside the tube.
Applying Gauss’s theorem

qin 1
φnet = =0 So Ein = 0 Eout ∝
εo r
Ein = 0
r
r =R

11.5 E due to infinitely long solid cylinder of radius R having uniformly distributed charge in
volume (charge density ρ)

(A) E at outside point
Let’s choose a cylindrical Gaussian surface. ∞

Applying Gauss's law r

q ρ × πR2 
E × 2πrλ = in =
εo εo

ρR2 1
⇒ Eout = ∝
2r εo r

∞
 
(B) E at inside point
Let’s choose a cylindrical Gaussian surface inside the solid cylinder. Applying Gauss's law
∞

qin 2
E × 2πr = = ρ × πr  r
εo εo
ρr
⇒ Ein = ∝r 
2εo

∞
Ein ∝ r 1
Eout ∝
r

r
r =R
11.6 Electric field due to a conducting and non-conducting uniformly charge infinite sheets

Suppose Q charge
is given to

Conducting Non − conducting

plate plate
+ + +
+ + +
+
+ + +
+
+ + +

Electric field for both the cases

Q
E=
2Aε0

σconducting σnon −conducting

E= E=
εo 2εo
Q Q
where σconducting = where σnon −conducting =
2A A
Because Qis distributed Because Qis distributed
in' 2A 'area. in' A 'area.

Example 22: Q 2Q
Two large parallel conducting sheets (placed at finite distance) are given
charges Q and 2Q respectively. Find out charges appearing on all the
P Q
surfaces.

Solution:
Let there is x amount of charge on left side of first plate, so on
x Q–x y 2Q – y
its right side charge will be Q–x, similarly for second plate
E3 P Q
there is y charge on left side and 2Q – y charge is on right side
of second plate E 4 E2 E1

Ep = 0 (By property of conductor) 1 2 3 4

 x Q – x y 2Q – y 
⇒ –  + +

= 0
2Aεo  2Aεo 2Aεo
 2Aεo 
we can also say that charge on left side of
P = charge on right side of P
3Q –Q
x = Q – x + y + 2Q – y ⇒ x = ,Q–x=
2 2 +3Q –Q Q +3Q
2 2 2 2
Similarly, for point Q:
x + Q – x + y = 2Q – y ⇒ y = Q/2, 2Q – y = 3Q/2

So final charge distribution of plates is as shown in figure.

Concept Builder-4

Q.1 Two concentric rings, one of radius R and total charge +Q and the second of radius 2R and
total charge – 8 Q, lie in x – y plane (i.e., z = 0 plane). The common centre of rings lies at
origin and the common axis coincides with z-axis. The charge is uniformly distributed on both
rings. At what distance from origin is the net electric field on z-axis would be zero ?

z − axis
Where the
fieldis zero
R
z=0
2R
+Q

– 8Q

R R R
(1) (2) (3) (4) 2R
2 2 2 2

Q.2 An electron and a proton start from the negative plate and positive plate respectively, and go
to opposite plates. Which of them wins this race? (plates are parallel to each other)


Q.3 What is the electric flux linked with following geometries (E = E0ˆi)
sphere

E E = Eo ˆi
(A) (B)
R a b
c

(C) h (D) E = E0 ˆi
E0 ˆi h

R
  3  4  3
Q.4 The electric field in a region is given by
= E E0 i + E0 j with E0 = 2.0 × 10 N/C. Find the flux of
5 5
2
this field through a rectangular surface of area 0.2m parallel to the Y–Z plane.

Q.5 A cylinder of length L and radius b has its axis coincident with the x-axis. The electric field in

this region is E = 200iˆ . Find the flux through the left end of the cylinder.
2 2 2
(1) 0 (2) 200 πb (3) 100 πb (4) –200 πb

Q.6 region of space the electric field is E = E0 xiˆ Consider an imaginary cubical volume of edge ‘a’
with its edges parallel to the axes of coordinates. The charge inside this volume will be
3 1 1
(1) zero (2) ε0E0 a (3) E a3 (4) ε E a3
ε0 0 6 0 0
Q.7 Calculate flux through cylinder ?

 = 1m

(1) λ·ε0 (2) λ2 ε0 (3) λ / ε0 (4) ε0 / λ

Q.8 S1 and S2 are two hollow concentric spheres enclosing charges Q and 2Q, 2Q
respectively, as shown in figure. Q
(i) What is the ratio of the electric flux through S1 and S2
(ii) How will the electric flux through sphere S1 change if medium S1
S2
of dielectric constant 5 is introduced in the space inside S1 in place of air.

Q.9 Figure shows three large metallic plates with charges Q, 4Q and –Q respectively. Determine
the final charges on all the surfaces.
Q 4Q –Q

1 2 3

12. Electrostatic Potential Energy

12.1 Electrostatic potential energy of a point charge due to many charges
The electrostatic potential energy of a point charge at a point in q1 Source charges
electric field is the work done in taking the charge from reference Test charges
q2
point (generally at infinity) to that point without acceleration (or q3 q
keeping KE const. or Ki = Kf). Its Mathematical formula is
q4
U = W(∞→P)ext = qV = – W(P→∞)ele.
Here q is the charge whose potential energy is being calculated and V is the potential at its
position due to the source charges.

Note: Always put q and V with sign.

Properties
(i) Electric potential energy is a scalar quantity but may be positive, negative or zero.
(ii) Its unit is same as unit of work or energy that is joule (in S.Ι. system). Some times energy is
also given in electron-volts.
–19
1eV = 1.6 × 10 J
(iii) Electric potential energy depends on reference point. (Generally Potential Energy at r= ∞ is
taken zero)

Example 23:
The four identical charges q each are placed at the corners of a square of side a. Find the
potential energy of one of the charges due to the remaining charges.
Solution:
The electric potential energy of point A due to the charges placed at B, C and D is
q a
1 q2 1 q2 1 q2 q
UA = + + D C
4πε0 a 4πε0 2a 4πε0 a
a a
2
1  1 q
= 2 +  qA Bq
4πε0  2 a a

13. Electrostatic Potential Energy of A System of Charges

(This concept is useful when more than one charges move)
It is the work done by an external agent against the internal electric field required to make a
system of charges in a particular configuration from infinite separation without accelerating it.
13.1 Types of system of charge
(i) Point charge system
(ii) Continuous charge system

Example 24:
Q
If electric potential energy of given system is positive then prove that 2Q > 3q
Solution: a a
U(system) = Sum of potential energy of all pairs
k(Q)(2Q) k2Qq kQq kQ
= – – = (2Q – 3q) 2Q a –q
a a a a
Given U is positive means U > 0
kQ
⇒ (2Q – 3q) > 0] ⇒ 2Q > 3q
a

Example 25: –Q
Figure shows an arrangement of three point charges. The total +q +q
r
q
potential energy of this arrangement is zero. Calculate the ratio . 2r
Q
Solution:
1  –qQ (+q)(+q) Q(–q) 
Usys =  + + =0
4πε0  r 2r r 
q q q 4
–Q + –Q=0 ⇒ 2Q = ⇒ =
2 2 Q 1
Example 26:
Two point charges each of mass m and charge q are released when they are at a distance r
from each other. What is the speed of each charge particle when they are at a distance 2r?
Solution:
According to momentum conservation both the charge particles will move with same speed
now applying energy conservation.
V V

2r

Kq2 1  Kq2 Kq2

k1 + k2 + ui = uf + kf ⇒0 + 0 + = 2  mv 2  + ⇒v=
r 2  2r 2rm

Example 27:
A q
Three equal charges q are placed at the corners of an equilateral triangle
of side a.
(i) Find out potential energy of charge system.
(ii) Calculate work required to decrease the side of triangle to a/2. B C
Solution: q q
(i) U = U12 + U13 + U23 a
2
Kq Kq 2
Kq 2
3Kq2
⇒ + + =
a a a a

(ii) Work required to decrease the sides to a/2

W = Uf – U i
3Kq2 3Kq2 3Kq2
⇒ – =
a/2 a a

14. Electric Potential

In electrostatic field the electric potential (due to some point source charges) at a point P is
defined as the work done by external agent in taking a unit positive point charge from a
reference point (generally taken at infinity) to that point P without changing its kinetic energy.
OR
Electric potential at a point is also equal to the negative of the work done by the electric field
in taking the point charge from reference point (i.e. infinity) to that point.
Mathematical representation
If (W ∞ → P)ext is the work required in moving a point charge q from infinity to a point P, the
electric potential of the point P is
(W∞→P )ext  (–Wele )∞→p ∆U Up – U∞ Up
Vp =  = = = =
q  ∆K =0 q q q q
(W∝ → P)ext can also be called as the work done by external agent against the electric force on a
unit positive charge due to the source charge.

14.1 Properties
(i) Potential is a scalar quantity, its value may be positive, negative or zero.
joule 1 2 –3 –1
(ii) S.Ι. Unit of potential is volt = and its dimensional formula is [M L T Ι ].
coulomb
(iii) Electric potential due to a positive charge is always positive and due to negative charge it
is always negative except at infinite. (taking V∞=0)
 (iv) Potential decreases in the direction of electric field.
(v) V = V1 + V2 + V3 + .......
(vi) Potential at reference point may or may not be zero.
14.2 Use of potential
If we know the potential at some point (in terms of numerical value or in terms of formula)
then we can find out the work done by electric force when charge moves from point 'P' to ∞
by the formula ( Wele )p →∞ = qVp

Example 28:
A charge 2µC is taken from infinity to a point in an electric field, without changing its velocity.
If work done against electrostatic forces is –40µJ then find the potential at that point.
Solution:
Wext –40µJ
V= = = –20 V
q 2µC
Example 29:
When charge 10 µC is shifted from infinity to a point in an electric field, it is found that work
done by electrostatic forces is 10 µJ. If the charge is doubled and taken again from infinity to
the same point without accelerating it, then find the amount of work done by electric field
and against electric field.
Solution:
(Wext)∞→p = (–wele)∞→p = (wele)p→∞ = 10 µJ
because ∆KE = 0
(Wext )∞→p
10µJ
Vp = = 1V =
q 10µC
So if now the charge is doubled and taken from infinity then
(Wext )∞→p
1= ⇒ (Wext)∞→P = 20 µJ ⇒ (Wele)∞→P = –20 µJ
20µC

Example 30:
A charge 3µC is released at rest from a point P where electric potential is 20 V then its kinetic
energy when it reaches to infinite is:
Solution:
Wele = ∆K = Kf – 0 (Wele)P → ∞ = qVP = 60 µJ So, Kf = 60 µJ

14.3 Potential due to a point charge

r  
–∫ (q0E)·∆ r
Wext( ∞→p) r 

V= = ∞ = –∫ E · dr r
qo qo ∞
Q P
r
KQ KQ
⇒ V = –∫ (–dr)cos 180º =
∞ r 2
r

Example 31:
Four point charges are placed at the corners of a square of side  calculate potential at the
centre of square.
+Q –3Q
Solution:
k ( Q-3Q-2Q+4Q )
=VC = 0 C
/ 2
–2Q +4Q
14.4 Potential due to a ring
(A) Potential at the centre of uniformly or non uniformly charged ring
Potential due to the small element dq
Kdq Kdq q,R
dV =
R
; Net potential V= ∫ R
dq

K Kq R
V= ∫ dq =
R R

(B) Potential due to half ring at center is

Kq
V= R
R
+q

(C) Potential at the axis of a ring

Kq dq
V= R
R2 + x 2

R2 + x2
x P
q

14.5 Potential Due To Uniformly Charged Spherical shell

As we know
r= r
 
V= –
r →∞
∫ E · dr

(A) At outside point (r > R):

r= r
 KQ  KQ
Vout = – ∫  2  dr ⇒ Vout =
r →∞  r  r
For outside point, the hollow sphere act like a point charge.
(B) Potential at inside point ( r < R )
Suppose we want to find potential at point P, inside +Q,R
+ +
the sphere. ++ +
+ P +
Potential difference between Point P and O: + +
+ r +

P
 + O +
VP - VO = –∫ Ein · dr Where Ein = 0 + +
O + +
+ + +
KQ + +
So VP - V O = 0 ⇒ VP = VO =
R
KQ KQ
⇒ VIN = =
R (Radius of the sphere)
14.6 Potential Due To Uniformly Charged Solid Non Conducting Sphere
KQ
• for r ≥ R (outside) V=
r dx
• for r ≤ R (inside) x

V=
KQ 2
(3R – r ) =
2 (
ρ 3R – r 2 2
)
2R 3
6ε0
Q
Here ρ=
4 3
πR
3

Example 32:
Two concentric spherical shells of radius R1 and R2 (R2 > R1) are having uniformly distributed
charges Q1 and Q2 respectively. Find out potential C
(i) at centre point A R2 B
(ii) at surface of smaller shell (i.e. at point B)
(iii) at surface of larger shell (i.e. at point C) A R
1
Q2
(iv) at r ≤ R1 Q 1

(v) at R1 ≤ r ≤ R2
(vi) at r > R2
Solution:
According to formula
KQ1 KQ2 KQ1 KQ2
(i) VA = + (ii) VB = +
R1 R2 R1 R2
KQ1 KQ2 KQ1 KQ2
(iii) VC = + (iv) for r ≤ R1 V = +
R2 R2 R1 R2
KQ1 KQ2 KQ1 KQ2
(v) for R1 ≤ r ≤ R2 V = + (vi) for r > R2 V = +
r R2 r r

Example 33:
Two hollow concentric nonconducting spheres of radius a and b (a > b) contains charges
Qa and Qb respectively. Prove that potential difference between two spheres is independent of
charge on outer sphere. If outer sphere is given an extra charge, is there any change in
potential difference ?
Solution:
KQb KQa
Vinner sphere = + Qa
b a Qb
KQb KQa b a
Vouter sphere = +
a a
KQb KQb
Vinner sphere – Vouter sphere = –
b a
 1 1
∆V = KQb  – 
b a 
Which is independent of charge on outer sphere. If outer sphere is given any extra charge then
there will be no change in potential difference.
 Concept Builder-5
–5
Q.1 Two charged particles each having equal charges 2 × 10 C are brought from infinity to within
a separation of 10 cm. Calculate the increase in potential energy during the process and the
work required for this purpose.
Q.2 Plot the following graphs –
(i) Electric field inside a conducting sphere with distance from centre
1
(ii) E versus where E is electric field due to a point charge and r is the distance from charge
r
(iii) Electric potential energy (U) of a pair of 2 like charges with distance (r) between charges.

Q.3 A particle A has charge +q and particle B has charge +4q with each of them having the same
mass m. When allowed to fall from rest through the same electric potential difference, the
V
ratio of their speeds A will become
VB
(1) 1: 2 (2) 2: 1 (3) 1: 4 (4) 4: 1

Q.4 What is the potential at the centre of the triangle?

+2Q

a a

+Q a –3Q

Q.5 Find the potential at origin ?

(0, +a) 1C
(–2a,0) (a,0) (4a,0)
4C –2C +3C
(0,–a) –3C

Q.6 A point charge q0 having mass m is placed at the centre of uniformly charged ring of total
charge Q and radius R. If the point charge is slightly displaced with negligible force along axis
of the ring then find out its speed when it reaches to a large distance.
Q.7 Four point charges –Q, –q, 2q and 2Q are placed, one at each corner of the square. The
relation between Q and q for which the potential at the centre of the square is zero is:
(1) Q = –q (2) Q = – 1/q (3) Q = q (4) Q = 1/q

Q.8 Two point charges 10 µC and 5 µC are placed at points A and B A

respectively with AC = 40 cm. The work done by external force
in displacing the charge 5 µC form B to C, where BC = 30 cm
81 9 9
(1) 9 J (2) J (3) J (4) J
20 25 40 B
C

Q.9 If an electron moves from rest from a point at which potential is 50 volt to another point at
which potential is 70 volt, then its kinetic energy in the final state will be
–10 –18
(1) 3.2 × 10 J (2) 3.2 × 10 J (3) 1 J (4) 1 dyne
Q.10 Charge on the outer sphere is q, and the inner sphere is grounded.
Then the charge on the inner sphere is q’, for (r2 > r1) r2
r1
(1) Zero (2) q’ = q
r1 r1
(3) q’ = q (4) q’ = – q
r2 r2

15. Potential Difference

The potential difference between two points A and B is work done by external agent against
electric field in taking a unit positive charge from B to A without acceleration (or keeping
Kinetic Energy constant or Ki = Kf).
• Mathematical representation
If (WA → B)ext = work done by external agent against electric field in taking the unit charge from
A to B
B  
–∫ Fe .dr
(WA →B )ext –(WA →B )electric UB – UA A
B  
V B – VA = = = = = –∫ E . dr
q ∆K =
0
q q q A

Note: Take W and q both with sign

15.1 Properties
(i) The difference of potential between two points is called potential difference. It is also called
voltage.
(ii) Potential difference is a scalar quantity. Its S.I. unit is volt.
(iii) If VA and VB be the potential of two points A and B, then work done by an external agent in
taking the charge q from A to B is
(Wext)AB= q (VB – VA) or (Wel) AB = q (VA – VB).
(iv) Potential difference between two points is independent of reference point.
15.2 Potential difference in a uniform electric field
 
VB – VA = – E · AB B

E
⇒ VB – VA = – |E| |AB| cos θ
θ
= – |E| d = – Ed A
d
d = effective distance between A and B along electric field. or we can
∆V
also say that E =
∆d
Special Cases :
Case 1. Line AB is parallel to electric field.
d
A B
E

∴ VA – VB = Ed
Case 2. Line AB is perpendicular to electric field.
A
d E
B

∴ VA – VB = 0 ⇒ VA = V B
Note: In the direction of electric field potential always decreases.
Example 34:
1µC charge is shifted from A to B and it is found that work done by an external force is 40µJ
in doing so against electrostatic forces then, find potential difference VA – VB
Solution:
(WAB)ext = q(VB – VA)
⇒ 40 µJ = 1µC (VB – VA)
⇒ VA – VB = – 40 V

Example 35:
A uniform electric field is present in the positive x-direction. Ιf the intensity of the field is
5N/C then find the potential difference (VB –VA) between two points A (0m, 2 m) and B (5 m, 3 m)
Solution:
 
VB – VA = – E . AB

= – (5 î ) · (5 î + ˆj ) = –25V
∆V
The electric field intensity in uniform electric field E =
∆d
Where ∆V = potential difference between two points.
∆d = effective distance between the two points. (projection of the displacement along the
direction of electric field)

Example 36: D
Find out following
uniform electric
(i) VA – VB (ii) VB – VC fieldE = 20N / C
C
(iii) VC – VA (iv) VD – VC 2cm
(v) VA – VD A B
2cm
(vi) Arrange the order of potential for points A, B, C and D.
Solution:
–2
(i) ∆VAB =
Ed = 20 × 2 × 10 = 0.4 So, VA – VB = 0.4 V
–2
(ii) ∆VBC =
Ed = 20 × 2 × 10 = 0.4 So, VB – VC = 0.4 V
–2
Ed = 20 × 4 × 10
(iii) ∆VCA = = 0.8 So, VC – VA = – 0.8 V

(iv) ∆VDC =
Ed = 20 × 0 = 0 So, VD – V C = 0
–2
(v) ∆VAD =
Ed = 20 × 4 × 10 = 0.8 So, VA – VD = 0.8 V
(vi) The order of potential VA > VB > VC = V D

16. Equipotential Surface

If potential of a surface (imaginary or physically existing) is same throughout then such
surface is known as an equipotential surface.

16.1 Properties of equipotential surfaces

(i) When a charge is shifted from one point to another point on an equipotential surface then
work done against electrostatic forces is zero.
(ii) Electric field is always perpendicular to equipotential surfaces.
(iii) Two equipotential surfaces do not cross each other.
16.2 Examples of equipotential surfaces
(A) Point charge: V2
V1
Equipotential surfaces are concentric and spherical as shown in figure. In
figure we can see that sphere of radius R1 has potential V1 throughout its q
R1
surface and similarly for other concentric sphere potential is same. R2

(B) Line charge:

V2
Equipotential surfaces have curved surfaces as that of coaxial cylinders
of different radii. V1

(C) Uniformly charged large conducting / non conducting sheets: +

V1 V2 V3

Equipotential surfaces are parallel planes. +

+
+
+
+
+
+

Note: In uniform electric field equipotential surfaces are always parallel planes.

Example 37:
Some equipotential surfaces are shown in figure. Calculate electric field with direction?
y(cm)
10V 20V 30V 40V

0 30º 30º 30º 30º

10 20 30 40 x(cm)

Solution:
Here we can say that the electric field will be perpendicular to equipotential surfaces
 ∆V
Also |E| =
∆d
where ∆V = potential difference between two equipotential surfaces
∆d = perpendicular distance between two equipotential surfaces
 10
So, E= = 200 V/m
(10 sin30º ) × 10–2
Now there are two perpendicular directions since we know that in the direction of electric
field electric potential decreases so the correct direction of electric field is making an angle
120° with the x-axis having value E = 200 V/m

Example 38:
Compare field, potential and surface charge density at A, B, and C.
Solution:
Surface of a metal is an equipotential surface (EPS) so charge on irregular shaped metal
distributes to create same potential on surface
⇒ VA = VB = VC Now radius of curvature (R) for straight line = ∞
 QA QB QC
RC > RB > RA Now = = or Q ∝ R
RA RB RC +
+ +
++ ++
+ ++ ++ +
+ B +
+ A ++
+
KQ V 1 ++ +
++ +++ +
Electric field E = ⇒E= ⇒E∝ ⇒ E A > EB > E C +
+
+
R2
R R + METAL
+
+
+ +
Q + C +
+++ + +
surface charge density σ ∝ ++ ++++ ++ + +
R2
V 1
⇒σ= ∝ ⇒ σA > σB > σ C
R R
So, if there is any possibility of charge leakage, it starts from point A. (corona discharge)
Remember that E and σ at sharp points is more while potential is same
Example 39:
Figure shows some equipotential surface produced by some charges. At which point the value
of electric field is greatest?

50V
40V
30V
20V
B

A C

Solution:
E is larger where equipotential surfaces are closer. ELOF are ⊥ to equipotential surfaces. In
the figure we can see that for point B they are closer so E at point B is maximum.

17. Relation Between Electric Field Intensity And Electric Potential

(A) For uniform electric field

(i) Potential difference between two points A and B B
 →
VB – VA = – E. AB .
A
(B) Non uniform electric field
∂V ∂V ∂V
(i) Ex = – , Ey = – , Ez = –
∂x ∂y ∂z
  ∂v ˆ ∂v ˆ ∂v 
⇒ E = Ex î + Ey ˆj + Ez k̂ = – ˆi +j +k 
 ∂x ∂y ∂z 
∂V
Where = derivative of V with respect to x (keeping y and z constant)
∂x
∂V
= derivative of V with respect to y (keeping z and x constant)
∂y
∂V
= derivative of V with respect to z (keeping x and y constant)
∂z

(C) If electric potential and electric field depends only on one coordinate, say r
rB
 ∂V    
(i) E = – r̂ (ii) ∫ dV = – ∫ E · dr ⇒ VB – VA = – ∫ E . dr
∂r rA
r  
(iii) The potential of a point ⇒ V = – ∫ E . dr
∞
Example 40:
The electric potential in a region along x-axis varies with x according to the relation
2
V(x) = 4 + 5x . Then the incorrect statement is-
(1) potential difference between the points x = 1 and x = 2 is 15 volt
(2) force experienced by a one coulomb charge at x = –1 m will be 10 N
(3) the force experienced by the above charge will be towards + x-axis
(4) a uniform electric field exists in this region along the x-axis
Solution:
2
V(x) = 4 + 5x
2 2
V(1) = 4 +5(1) = 9 volt; v(2) = 4 + 5(2) = 24 volt
Potential difference = (24 – 9) volt = 15 volt
dV d 2
E=– =– [4 + 5x ] = –10x
dx dx
EF is variable, depends on x
–1
Electric field at x = –1m is |E| = 10 NC
So, force on 1C charge is 10N. It is towards +x axis.
So, statement (4) is incorrect.

Example 41:
A uniform electric field is along x–axis. The potential difference VA– VB = 10 V between two
points A (2m,3m) and B (4m, 3m). Find the electric field intensity.
Solution:
∆V 10
E= = =5V/m ⇒ It is along + ve x-axis.
∆d 5

Example 42:
2

V = x + y , Find E
Solution:
∂V ∂V ∂V
= 2x, = 1 and =0
∂x ∂y ∂z
  ∂V ˆ ∂V ˆ ∂V 
E = –  ˆi +j +k  = –(2x î + ˆj ) Electric field is nonuniform.
 ∂x ∂y ∂z 

Concept Builder-6

Q.1 Two parallel plates separated by a distance of 5 mm are kept at a potential difference of 50
–15 –11 7
V. A particle of mass 10 kg and charge 10 C enters in it with a velocity 10 m/s. The
acceleration of the particle will be
8 2 5 2 5 2 3 2
(1) 10 m/s (2) 5 × 10 m/s (3) 10 m/s (4) 2 × 10 m/s

Q.2 In moving from A to B along an electric field line, the electric field does
6.4 × 10
–19
J of work on an electron. If φ1, φ2 are equipotential surface, B
E
then the potential difference (VC – VA) is A C
(1) –4 V (2) 4 V φ1
φ2
(3) Zero (4) 64 V
 2
Q.3 The electric potential V at any point O (x, y, z all in metres) in space is given by V = 4x volt.
The electric field at the point (1 m, 0, 2 m) in volt/meter is
(1) 8 along negative X-axis
(2) 8 along positive X-axis
(3) 16 along negative X-axis
(4) 16 along positive Z-axis


Q.4 E (yiˆ + xj)
The potential of an electric field = ˆ is
(1) V = –xy + constant (2) V = –(x + y) + constant
2 2
(3) V = –(x + y ) + constant (4) V = constant


Q.5 The electric field intensity at all points in space in given by E = 3 ˆi – ˆj volts/meter. The nature
of equipotential lines in x – y plane is given by
Highpotential Highpotential
Low potential Low potential
y y
y y

30º 60º
x 30º x 60º
x x
(1) (2) (3) (4)

Low potential Highpotential Low potential Highpotential

18. Electric Dipole

In some molecules, the centres of (+ve) and (-ve) charge do not coincide. This results in the
formation of electric dipole. Atom is non - polar because in the centre of atom (+ve) and (-ve)
charges coincide. Polarity can be induced in an atom by the application of electric field. Hence
it can be called as induced dipole.

18.1 Dipole Moment

 
Dipole moment P = q x 2   is of Å order
2
(i) Vector quantity directed from (-ve) charge to (+ve) charge
+ –
(ii) Dimension: [LTA] +q  –q
p
Unit: coulomb × meter (or Cm)
(iii) Practical unit is debye
–30
1 debye = 3.3 × 10 C×m

18.2 Electric field at general point (r, θ): polar coordinates

K(P sin θ)

Er = E
KP r3 2K(Pcos θ)
Er =
Enet = 3
1 + 3cos2 θ r3
r α
θ 
tan θ P
tan α = ; r A
2
Pcos θ
kpcos θ 1
Vp = ∝ θ +q
r2 r2 –q

P sinθ
 End on or Broad on or
Axial position Equatorial
θ =0º position θ =90º
 P
 E
p 
E
– +
O P r
r 90º
O
+  –
p
1 p
VP = VP = 0
4π ∈0 r 2
  
  1   2p   1   –p 
EP =  EP =   3 
 3   4πε0   r 
 4πε0   r 

18.3 Dipole Placed in uniform Electric Field

 
p = 2q 
  
(A) Fnet = qE + (–q)E = 0

(B) Torque

+ qE

p θ  
E p
θ 
E
–   
qE τ(= p × E)
       
τ = r × F = 2 × qE = 2q  × E = p × E

(C) Work: Work done in rotating an electric dipole from θ1 to θ2 [in uniform field]
dW = τ dθ ⇒W = ∫ dW = ∫ τdθ
 + 
θ2 
E p E
Wθ
1 →θ2
= ∫ pE sinθ dθ = pE (cosθ1 – cosθ2) –  +

+  –
θ1
p E p
W0 →180 = pE [1– (–1)] = 2 pE –
θ =0 θ =90º θ =180º
W0 → 90 = pE (1–0) = pE τ =0(min) τ =pE(max) τ =0(min)
If a dipole is rotated from field direction W = 0(min) W = pE W = 2pE(max)
(θ = 0°) to θ then W = pE ( 1- cosθ)
Electrostatic potential energy of a dipole placed in a uniform field is defined as work done in
rotating a dipole from a direction perpendicular to the field to the given direction i.e.
U = Wθ – W90 = pE (1–cos θ ) – pE
 
= – pE cos θ = – p.E

E is a conservative field so whatever work is done in rotating a dipole from θ1 to θ2 is just
equal to change in electrostatic potential energy.
Wθ = Uθ – Uθ = pE (cos θ1– cos θ2)
1 →θ2 2 1
 Angular SHM: When a dipole suspended in a uniform field it will align itself parallel to the
field. Now if it is given a small angular displacement θ about its equilibrium position the
restoring torque will be.
τ = pE sinθ
If θ is small then τ = – pE θ

τ ∝ (– θ) (Angular S.H.M) q

But τ = I α p 
θ
E
So,
–q
Ια = – pEθ
pE 2
⇒α= (– θ ) = – ω θ (where α – angular acceleration & Ι – moment of Inertia)
Ι
pE 2π I
ω= ; T= = 2π
I ω pE

Example 43:
What is the dipole moment of the system shown in figure
–q

a a

2q a –q

Solution:

There are two diploes of P = q (a)

P
A
60º

P

so Pnet = 3p= 3 qa

Example 44:
–7 –7
A system has two charges qA = 2.5 × 10 C and qB = – 2.5 × 10 C located at points
A(0, 0, –0.15m) and B(0, 0, +0.15m) respectively. What is the net charge and electric dipole
moment of the system?
Solution:
–7 –7
Net charge = 2.5 × 10 – 2.5 × 10 = 0
Electric dipole moment
P = (Magnitude of charge) × (Separation between charges)
–7
= 2.5 × 10 [0.15 + 0.15] C m
–8
= 7.5 × 10 C m
The direction of dipole moment is from B to A.

Example 45:
The electric field due to a short dipole at a distance r, on the axial line, from its mid point is
the same as that of electric field at a distance r', on the equatorial line, from its
r
mid-point. Determine the ratio .
r´
Solution:
1 2p 1 p 2 1
= or 3 = 3
4πε0 r 3 4πε0 r '3 r r'
r3 r 1/3
or = 2 or =2
r'3
r'

Example 46:
Two charges, each of 5 µC but opposite in sign, are placed 4 cm apart. Calculate the electric
field intensity of a point that is at a distance 4 cm from the mid point on the axial line of the
dipole.
Solution:
We can not use formula of short dipole here because distance of the point is comparable to
the distance between the two point charges.
–6 –2 –2
q = 5 × 10 C, a = 4 ×10 m, r = 4 × 10 m 4cm
K(5µC) K(5µC) –5µC 5µC
Eres = E+ + E– = – P
(2cm)2 (6cm)2
4cm
144 –1 8 –1
= NC = 10 N C
144 × 10–8

Concept Builder-7

Q.1 Three charges of (+4Q), (–3Q) and (–Q) are placed at the corners A, B and C
+4Q
of an equilateral triangle of side a as shown in the adjoining figure.
Then the dipole moment of this combination is A
a a
Qa
(1) (2) zero
13 B C
a
2 –3Q –Q
(3) Qa 13 (4) Qa
13

Q.2 An electrical dipole of moment ‘p’ is placed in an electric field of intensity ‘E’. The dipole
acquired a position such that the axis of the dipole makes an angle θ with the direction of the
field. Assuming the potential energy of the dipole to be zero when θ = 90°, the torque and the
potential energy of the dipole will respectively be:
(1) pE sinθ, 2pE cosθ (2) pE cosθ, –pE sinθ
(3) pE sinθ, –pE cosθ (4)pE sinθ, –2pE cosθ

Q.3 A given charge is situated at a certain distance from an electric dipole in the end-on position
experiences a force F. If the distance of the charge is doubled, the force acting on the charge
will be
(1) 2F (2) F/2 (3) F/8 (4) 4F

Q.4 Two electric dipoles of moment P and 64 P are placed in opposite direction on a line at a
distance of 25 cm. The electric field will be zero at point between the dipoles whose distance
from the dipole of moment P is
25 4
(1) 5 cm (2) cm (3) 10 cm (4) cm
9 13
19. Conductor
19.1 Finding field due to a conductor

Suppose we have a conductor, and at any 'A', local + + +

+ +
surface charge density = σ. We have to find electric A E=?
+ +
field just outside the conductor surface. + +
+ +
+ +
For this lets consider a small cylindrical Gaussian +
+
+
+ + +
surface, which is partly inside and partly outside the + +

conductor surface, as shown in figure. It has a small

cross section area ds and negligible height. (2)ds
(3) E
Ein = 0 (1)
Cylindrical
gaussian
surface

Applying Gauss's theorem for this surface

σds σ qin σds
So, Eds = ⇒E= φnet = =
ε0 ε0 ε0 ε0

Electric field just outside the surface of flux through flux through flux through
σ surface (1) surface(2) surface(3)
conductor E = direction will be normal f1 = Eds  f2 = 0 f3 = 0
ε0 (because E is (Eisnormal (as E inside
 σ normal to tocurved the conductor= 0)
to the surface. In vector form E = n ˆ the surfaceof Gaussiansurface)
ε0 conductor)
(here = n̂ unit vector normal to the conductor surface)

19.2 Conductor and it's properties [For electrostatic condition]

(i) Conductors are materials which contains large number of free electrons which can move
freely inside the conductor.
(ii) Ιn electrostatics, conductors are always equipotential surface
(iii) Charge always resides on outer surface of conductor.
(iv) Ιf there is a cavity inside the conductor having no charge then charge will always reside
only on outer surface of conductor.
(v) Electric field is always perpendicular to conducting surface.
(vi) Electric lines of force never enter into conductors.
(vii) Electric field intensity near the conducting surface is given by formula
 σ
E= n ˆ A
ε0
 σ
ˆ
EA = A n
ε0 C
B
 σ  σ
ˆ and EC = C n
EB = B n ˆ
ε0 ε0
(viii) When a conductor is grounded its potential becomes zero.
V=0
 (ix) When an isolated conductor is grounded then its charge becomes zero.
(x) When two conductors are connected there will be charge flow till their potential becomes
equal.
(xi) Electric pressure: Electric pressure at the surface of a conductor is given by formula
σ2
P= where σ is the local surface charge density.
2ε0

20. Sharing Of Charges

Two conducting hollow spherical shells of radii R1 and R2 having q1
R2
q2
R1
charges Q1 and Q2 respectively and separated by large distance, are
joined by a conducting wire
Let final charges on spheres are q1 and q2 respectively.

Potential on both spherical shell become equal after joining, therefore

Kq1 Kq2
=
R1 R2
q1 R1
⇒ = ....(i)
q2 R2
and q1 + q2 = Q1 + Q2 ....(ii)
from (i) and (ii)
(Q + Q2 )R1 (Q1 + Q2 )R2
q1 = 1 ; q2 = ;
R1 + R2 R1 + R2
q1 R1 σ 1 4πR21 R1
ratio of charges = ⇒ =
q2 R2 σ2 4πR 2
2
R2
σ1 R2
Ratio of surface charge densities =
σ2 R1
q1 R1
Ratio of final charges =
q2 R2

3Q
Example 47:
–Q
Two conducting hollow spherical shells of radii R and 2R carry charges –Q
and 3Q respectively. How much charge will flow into the earth if inner shell R
is grounded?
Solution: 2R
When inner shell is grounded to the Earth then the potential of inner shell
will become zero because potential of the Earth is taken to be zero.
Kx K3Q –3Q
+ = 0, x=
R 2R 2 3Q
the charge that has increased
x
–3Q –Q
= – (–Q) =
2 2 R
Q
Hence charge flows into the Earth =
2 2R
21. Mixing Of Identical Charged Tiny Drops
Let, number of tiny drops = N
[For each tiny drop] [For Big drop]
(r, q, σ, E,V) (R, Q, σB, EB, VB)

R
Ntiny drops Bigdrops

(i) Charge conservation Q = Nq

4 3 4 3
(ii) Volume conservation N πr = πR
3 3
1/3
Hence R = N r , Q = Nq
1/3 1/3 2/3
∴ σB = N σ , EB = N E, VB = N V

Example 48:
216 identical drops each charged to the same potential of 50 volts are combined to form a
single drop. The potential of the new drop will be
Solution:
2/3
V=n ×v
2/3
V = (216) × 50 = 36 × 50 = 1800 V

22. Van De Graph Generator

This is a machine that can build up high voltages of the order of a few million volts. The
resulting large electric fields are used to accelerate charged particles like electrons, protons,
ions etc.
++ + + +
+ + Pulley
Metalbrush++ +
+
+ +
+ +
++ + Insultingbelt
+ +
+ +
+ + + to carry and
+ + + deliver charge
+
+
+ Insulating
+ supporting
+ column
+
+ Motor driven
++ pully
+++
Metalbrush Grounded
deliveringcharge metalbase
from source

Construction and working

It is a machine capable of building up potential difference of a few million volts, and fields
6
close to the breakdown field of air which is about 3 × 10 V/m. A large spherical conducting
shell (of few metres radius) is supported at a height several meters above the ground on an
insulating column. A long narrow endless belt of insulating material like rubber or silk, is
wound around two pulleys- one at ground level, one at the centre of the shell. This belt is
kept continuously moving by a motor driving the lower pulley. It continuously carries positive
charge, sprayed on to it by a brush at ground level, to the top. There it transfers its positive
charge to another conducting brush connected to the large shell. Thus, positive charge is
transferred to the shell, where it spreads out uniformly on the outer surface. In this way
voltage differences of as much as 6 or 8 million volts (with respect to ground) can be built up.
 ANSWER KEY FOR CONCEPT BUILDERS

CONCEPT BUILDER-1 CONCEPT BUILDER-5

1. (4) 2. (2) 1. Work done ⇒ 36 Joule
2. –40 N (attraction) 3. 351 N ∆U = 36 joule

4. (2) 5. (4) E

2. (i) Econd = 0
CONCEPT BUILDER-2
1. (1) 2. (1) 3. (3) 
r
4. (1) 5. (2)
E

CONCEPT BUILDER-3 kQ 1 2 2
(ii) E = ; = x; E = KQx ∝ x
1. (3) 2. (3) 3. (1) r 2
r
4. (1) 5. a = 2g x

6.

Ep =
ˆ
9 × 103 –iˆ – 6ˆj – k (
N/C
)
U
38 38

7. = (
E21 3 2 ˆi + 7ˆj N/C ) (iii) U =
KQ1Q2
r
∝
1
r
8. (1) 9. (3) r
–q –q 3. (1) 4. vc = 0
10. (i) Q = ; (ii) Q =
3 4 45
5. – × 109 V
4a
CONCEPT BUILDER-4
2kQq0
1. (4) 6. v= m/s
mR
2. Electron wins the race
7. (1) 8. (4)
3. (A) φE = 0 (B) φE = 0
9. (2) 10. (4)
(C) φE = 0 (D) φE = 0
2
4. 240 Nm /C 5. (4) CONCEPT BUILDER-6
6. (2) 7. (3) 1. (1) 2. (2) 3. (1)
φ1 1 Q 4. (1) 5. (3)
8. (i) = (ii) φ1 =
φ2 3 5ε0
Q 4Q –Q
CONCEPT BUILDER-7
1. (3) 2. (3) 3. (3)
4. (1)
2Q –Q Q 3Q –3Q 2Q
9.
 Exercise - I

Charge and Properties 7. A charge Q is divided in two parts Q1

and Q2 and these charges are placed at
1. A body has 80 microcoulomb of charge.
a distance R. There will be maximum
Number of additional electrons on it
repulsion between them when
will be:
Q–Q Q
(1) 8 × 10
–5
(2) 80 × 10
15
(1) Q1
= = ,Q2
14 –17
R R
(3) 5 × 10 (4) 1.28 × 10
2Q Q
(2) Q1
= = , Q2
3 3
2. Which of the following charge can not
3Q Q
present on oil drop in Millikan's (3) Q1
= = ,Q2
4 4
experiment:–
–19 –19 Q
(1) 4.0 × 10 C (2) 6.0 × 10 C (4) Q1 = Q2 =
–19
2
(3) 10.0 × 10 C (4) all of them
8. The force of repulsion between two
3. What equal charges would to be placed point charges is F, when these are at a
on earth and moon to neutralize their distance of 0.1 m apart. Now the point
charges are replaced by spheres of
gravitational attraction (Use mass of
25 23
diameter 5cm having the charge same
earth = 10 kg, mass of moon = 10 kg) as that of point charges. The distance
13 26
(1) 8.6 × 10 C (2) 6.8 × 10 C between their centre is 0.1 m, then the
(3) 8.6 × 10 C
3 6
(4) 9 × 10 C force of repulsion will:
(1) increases (2) decreases
4. If in Millikan's oil drop experiment (3) remains same (4) 4 F
charges on drops are found to be 8µC,
9. Two charges 4q and q are placed at a
12µC, 20µC, then quanta of charge is:
distance  apart. An another charged
(1) 8µC (2) 4µC
(3) 20µC (4) 12µC particle Q is placed in between them
(at mid point). If resultant force on q is
zero then the value of Q is:
Coulomb's Law and Super Position of
(1) q (2) – q
Forces (3) 2q (4) – 2q
5. Coulomb’s law for the force between
electric charges most closely resembles 10. A total charge of 20µC is divided into two
parts and placed at some distance apart.
with:
If the charges experience maximum
(1) Law of conservation of energy
coulombian repulsion, the charges should
(2) Newton’s law of gravitation
be:
(3) Newton’s 2nd law of motion (1) 5µC, 15µC (2) 10µC, 10µC
(4) The law of conservation of charge
40 20
(3) 12µC, 8µC (4) µC, µC
6. A point charge q1 exerts a force F upon 3 3
another point charge q2. If a third
11. Two point charges placed at a distance
charge q3 be placed quite near the r in air exert a force F on each other.
charge q2 then the force that charge q1 The value of distance R at which they
exerts on the charge q2 will be: experience force 4F when placed in a
medium of dielectric constant 16 is:
(1) F (2) > F
(1) r (2) r/4
(3) < F (4) zero (3) r/8 (4) 2r
12. Two point charges of + 2 µC and + 6µC Electric Field Due to Point Charges, Null
repel each other with a force of 12 N. If Point and Field Line
each is given an additional charge of –
17. If an electron is placed in a uniform
4 µC, then force will become:
electric field, then the electron will:
(1) 4N (attractive)
(1) experience no force.
(2) 60 N (attractive)
(2) move with constant velocity in the
(3) 4 N (Repulsive)
direction of the field.
(4) 12 N (attractive)
(3) move with constant velocity in the
direction opposite to the field.
13. A negative charge is placed at some (4) accelerate in direction opposite to
point on the line joining the two +Q field.
charges at rest. The direction of motion
of negative charge will depend upon the: 18. The electric field in a certain region is
(1) position of negative charge alone.  K
(2) magnitude of negative charge alone. given by E = ( 3 ) ˆi . The dimensions of K
x
(3) both on the magnitude and position
are:
of negative charge. –3 –1
(1) MLT A
(4) magnitude of positive charge. –2 –3 –1
(2) ML T A
4 –3 –1
14. Two point charge q1 and q2 are placed (3) ML T A
(4) dimensionless charge
at a distance of 50 cm from each other
in air, and interact with a certain force.
19. Two point charges a & b, whose
Now the same charges are put in an oil
magnitudes are same are positioned at a
whose relative permittivity is 5. If the
certain distance from each other with a
interacting force between them is still
at origin. Graph is drawn between
the same, their separation now is: electric field strength at points between
(1) 16.6 cm (2) 22.3 cm a & b and distance x from a. E is taken
(3) 35.0 cm (4) 28.4 cm positive if it is along the line joining from
a to b. From the graph, it can be
15. Force between two identical spheres decided that
charged with same charge is F. If 50% E
charge of one sphere is transferred to
x
second sphere then new force will be: a b
3 3
(1) F (2) F
4 8 (1) a is positive, b is negative
3 (2) a and b both are positive
(3) F (4) none
2 (3) a and b both are negative
(4) a is negative, b is positive
16. Two balls carrying charges +7µC and
–5µC attract each other with a force F. 20. Two charges are placed as shown in fig.
If a charge –2µC is added to both, the Where should be a third charge be
force between them will be: placed so that it remains in rest
condition:–
F
(1) F (2) 9e 16e
2
70cm
(3) 2F (4) zero
(1) 30cm from 9e (2) 40cm from 16e
(3) 40cm from 9e (4) (1) or (2)
21. Charge 2Q and –Q are placed as shown 26. There is a uniform electric field in x-
in figure. The point at which electric direction. If the work done by external
field intensity is zero will be: agent in moving a charge of 0.2 C
–Q +2Q through a distance of 2 metre slowly
A B along the line making an angle of 60°
(1) Somewhere between –Q and 2Q. with x-direction is 4 joule, then the
(2) Somewhere on the left of –Q. magnitude of E is:
(3) Somewhere on the right of 2Q. (1) 3N/ C (2) 4 N/C
(4) Somewhere on the right bisector of (3) 5 N/C (4) 20 N/C
line joining –Q and 2Q.
27. The maximum electric field intensity on
22. The linear charge density on upper half the axis of a uniformly charged ring of
of a segment of ring is λ and at lower charge q and radius R will be:
half, it is –λ. The direction of electric 1 q 1 2q
field at centre O of ring is: (1) (2)
4πε0 3 3R2 4πε0 3R2
+ A
+
+ 1 2q 1 3q
+ (3) (4)
–
D O
B 4πε0 3 3R2 4πε0 2 3R2
–
–
–
C
Electric Flux, Density, Gauss's Law and
(1) along OA (2) along OB
Gaussian Surface, Application of Gauss's
(3) along OC (4) along OD
Law
Electric Field Due to Straight Wire, Sheet,
28. The total flux associated with given
Ring cube will be where 'a' is side of cube:
1 9
23. A ring of radius R is charged uniformly ( = 4π × 9 × 10 )
∈0
with a charge + Q. The electric field at
2µC
any point on its axis at a distance r a 1µC
from the circumference of the ring will 4µC 3µC
be:
6µC
KQ KQ 5µC
(1) (2) a
r r 2
8µC a 7µC
KQ KQr
(r )
1/2 4 2
(3) 2
–R 2
(4) (1) 162 π × 10 Nm /C
r3 R3 3 2
(2) 162 π × 10 Nm /C
24. Semicircular ring of radius 0.5 m. is 5
(3) 162 π × 10 Nm /C
2

uniformly charged with a total charge 6

(4) 162 π × 10 Nm /C
2
–9
of 1.4 × 10 C. The electric field
intensity at centre of this ring is: 29. A point charge is placed at a distance
(1) zero (2) 320 V/m. a/2 perpendicular to the plane and
(3) 64 V/m. (4) 32 V/m. above the centre of a square of side a.
The electric flux through the square is:
25. Two infinite linear charges are placed
parallel at 0.1 m apart. If each has q q
(1) (2)
charge density of 5µ C/m, then the ∈0 π ∈0
force per unit length of one of linear q q
charges in N/m is: (3) (4)
4 ∈0 6 ∈0
(1) 2.5 (2) 3.25
(3) 4.5 (4) 7.5
30. A charge q is placed at the centre of 34. Which of the following represents the
the cubical vessel (with one face open) correct graph for electric field intensity
as shown in figure. The flux of the and the distance r from the centre of a
electric field through the surface of the hollow charged metal sphere or solid
vessel is: metallic conductor of radius R:

E
d d E
(1) (2)
r
r
d

(1) zero (2) q/ε0

E E
q
(3) (4) 5q/6ε0
4ε0 (3) (4)
r r

31. Eight charges, 1µC, –7µC, – 4µC, 10µC,

35. A sphere of radius R and charge Q is
2µC, –5µC, -3µC and 6µC are situated
placed inside an imaginary sphere of
at the eight corners of a cube of side radius 2R whose centre coincides with
20 cm. A spherical surface of radius 80 the given sphere. The flux related to
cm encloses this cube. The centre of imaginary sphere is:–
the sphere coincides with the centre of
Q Q
the cube. Then the total outgoing flux (1) (2)
∈0 2 ∈0
from the spherical surface (in unit of
volt meter) is: 4Q 2Q
(3) (4)
(1) 36π × 10
3
∈0 ∈0
3
(2) 684π × 10
(3) zero 36. The intensity of an electric field at
(4) none of the above some point distant r from the axis of
infinite long pipe having charges per
32. Which one of the following pattern of unit length as q will be:
2
electric line of force can't possible:– (1) proportional to r
3
(2) proportional to r
(1) (3) inversely proportional to r
2
(4) inversely proportional to r

37. A closed cylinder of radius R and length

L is placed in a uniform electric field E,
(2)
parallel to the axis of the cylinder. Then
the electric flux through the cylinder
must be -
(3) (1) 2πR E
2 2
(2) (2πR + 2πRL)E
(3) 2πRLE (4) zero
(4)
38. In a region of space the electric field is

33. Total charge on a sphere of radii 10 cm given by E = 8 ˆi + 4ˆj + 3kˆ . The electric
is 1 µC. The maximum electric field due flux through a surface of area of 100
to the sphere in N/C will be: units in x–y plane is:
–5 3
(1) 9 × 10 (2) 9 × 10 (1) 800 units (2) 300 units
5 15
(3) 9 × 10 (4) 9 × 10 (3) 400 units (4) 1500 units
 Electrostatics Potential Energy Electric Potential due to a Point
39. Which statement is true: Charge, Ring and Infinite Long Wire
(i) A ring of radius R carries a 44. The dimension of potential difference
uniformly distributed charge +Q. A are:
2 –2 –1 –2 –1
point charge –q is placed on the (1) ML T C (2) MLT C
–2 –2 2 –1 –1
axis of the ring at a distance 2R (3) MT C (4) ML T C
from the centre of the ring and
released from rest. The particle 45. An object is charged with positive
executes a simple harmonic motion charge. The potential at that object will
be:
along the axis of the ring. (1) positive only
(ii) Electrons move from a region of (2) negative only
higher potential to that of lower (3) zero always
potential. (4) may be positive, negative or zero.
(1) only (i) (2) only (ii)
(3) (i), (ii) (4) none of them 46. Two points (0, a) and (0, – a) have
charges q and –q respectively then the
40. As shown in figure, on bringing a charge electrical potential at origin will be-
Q from point A to B and from B to C, (1) zero (2) kq/a
2
the work done are 2 joule and – 3 joule (3) kq/2a (4) kq/4a
respectively. The work done in bringing
47. The charges of same magnitude q are
the charge from C to A will be: placed at four corners of a square of
C side a. The value of potential at the
centre of square will be:
A B (1) 4kq/a (2) 4 2kq / a
(1) – 1 joule (2) 1 joule (3) 4kq 2a (4) kq / a 2
(3) 2 joule (4) 5 joule
48. A circle of radius R is drawn in a
41. In an electron gun, electrons are uniform electric field E as shown in the
accelerated through a potential fig. VA, VB, VC and VD are respectively the
difference of V volt. Taking electronic potential of point A, B, C and D at the
charge and mass to be respectively e periphery of the circle then
and m, the maximum velocity attained A
by them is:
D B
2eV 2eV
(1) (2)
m m
2 C
(3) 2 m/eV (4) (V /2em)
(1) VA > VC, VB = VD
42. If a charge is shifted from a low (2) VA < VC, VB = VD
potential region to high potential (3) VA = VC, VB < VD
region. The electrical potential energy: (4) VA = VC, VB > VD
(1) Increases
(2) Decreases 49. Figure represents a square carrying
(3) Remains constant charges +q, +q, –q, –q at its four
(4) May increase or decrease. corners as shown. Then the potential
will be zero at points:
43. You are given an arrangement of three +q P +q
point charges q, 2q and xq separated
by equal finite distances so that A C
electric potential energy of the system B
is zero. Then the value of x is:
2 1 –q Q –q
(1) – (2) –
3 3 (1) A, B, C, P and Q
(2) A, B and C
2 3 (3) A, P, C and Q
(3) (4)
3 2 (4) P, B and Q
 Potential due to sphere or shell, 55. In H atom, an electron is rotating
Potential Difference in a uniform around the proton in an orbit of radius
r. Work done by an electron in moving
Electric Field, Work Done in moving once around the proton along the orbit
Charge in Electric Field will be:
2 2
(1) ke/r (2) ke /r
50. Under the influence of charge, a point
(3) 2πre (4) zero
charge q is carried along different paths
from a point A to point B, then work 56. Choose the incorrect statement:
done will be: (1) the potential energy per unit
B positive charge in an electric field
at some point is called the electric
potential.
III (2) the work required to be done to
IV
II I move a point charge from one point
to another in an electric field
A depends on the position of the
(1) maximum for path four points.
(2) maximum for path one (3) the potential energy of the system
will increase if a positive charge is
(3) equal for all paths
moved against of Coulombian force.
(4) minimum for path three (4) the value of fundamental charge is
not equivalent to the electronic
51. A semicircular ring of radius 0.5 m is charge.
uniformly charged with a total charge
–9
of 1.5 × 10 . The electric potential at Relation between Electric Field and
the centre of this ring is: Potential Equipotential Surface
(1) 27 V (2) 13.5 V 57. The fig. shows lines of constant
(3) 54 V (4) 45.5 V potential in a region in which an
electric field is present. The value of
52. A particle of charge Q and mass m the potential are given. The magnitude
of the electric field is greatest at the
travels through a potential difference V
point:
from rest. The final momentum of the
A B
particle is:
mV
(1) (2) 2Q mV
Q C
50V
40V 30V
2QV 20V 10V
(3) 2mQV (4)
m (1) A (2) B
(3) C (4) A & C
53. A 5C charge experiences a force of 2000 58. Three equal charges are placed at the
N when moved between two points three corners of an equilateral triangle
separated by a distance of 2 cm in a as shown in the figure. The statement
uniform electric field. The potential which is true for electric potential V
difference between the two points is: and the field intensity E at the centre
(1) 8 volt (2) 80 volt of the triangle:
q
(3) 800 volt (4) 8000 volt

54. The work done to take an electron from

o
rest where potential is – 60 volt to
another point where potential is – 20
q q
volt is given by:
(1) 40 eV (2) – 40 eV (1) V = 0, E = 0 (2) V = 0, E ≠ 0
(3) 60 eV (4) – 60eV (3) V ≠ 0, E = 0 (4) V ≠ 0, E ≠ 0
59. Electric field at a distance x from origin Electric Dipole
100
is given as E = , then potential 65. For the given figure the direction of
x2
difference between points situated at electric field at A will be:
x = 10 m and x = 20 m: L X
(1) 5 V (2) 10 V
(3) 15 V (4) 4V A Y

60. The electric potential V is given as a Z

function of distance x (metre) by V = Q –Q
2 C
(5x –10x–9) volt. Value of electric field B
at x = 1 m is:– (1) towards AL (2) towards AY
(1) 20 V/m (2) 6 V/m (3) towards AX (4) towards AZ
(3) 11 V/m (4) zero

61. The variation of potential with distance 66. An electric dipole is placed in an
r from a fixed point is shown in Figure. electric field generated by a point
The electric field at r = 5 cm, is: charge:
(1) the net electric force on the dipole
5 must be zero.
(2) the net electric force on the dipole
V in volt

may be zero.
(3) the torque on the dipole due to the
0 2 4 6
r incm field must be zero.
(4) the torque on the dipole due to the
(1) (2.5) V/cm (2) (–2.5) V/cm
field may be zero.
(3) (–2/5) V/cm (4) (2/5) V/cm

62. The electric potential and electric field 67. The force on a charge situated on the
at any given position due to a point axis of a dipole is F. If the charge is
charge are 600 V and 200 N/C shifted to double the distance, the
respectively. Then magnitude of point acting force will be:
charge would be:
(1) 4F (2) F/2
(1) 3 µC (2) 30 µC
(3) F/4 (4) F/8
(3) 0.2 µC (4) 0.5 µC
63. The electric field in a region is directed 68. The electric potential at a point due to
outward and is proportional to the an electric dipole will be:
distance r from the origin. Taking the    
P. r P. r
electric potential at the origin to be (1) k 3 (2) k 2
zero, the electric potential at a r r
   
distance r: P×r P×r
(1) is uniform in the region (3) k 3 (4) k 2
r r
(2) is proportional to r
2
(3) is proportional to r 69. A small electric dipole is of dipole
(4) increases as one goes away from
moment p. The electric potential at a
the origin
distance 'r' from its centre and making
64. The electric field and the electric an angle θ from the axis of dipole will
potential at a point are E and V be:
respectively kp sin θ kpcos θ
(1) If E = 0, V must be zero. (1) 2
(2)
(2) If V = 0, E must be zero. r r2
(3) If E ≠ 0, V cannot be zero. kp kp
(3) 3
1 + 3cos2 θ (4) 1 + 3 sin2 θ
(4) None of these r r3
70. At any point on the right bisector of 76. For an electrostatic system which of
line joining two equal and opposite the statement is always true:
charges: (a) electric lines are parallel to metallic
(1) the electric field is zero. surface.
(2) the electric potential is zero. (b) electric field inside a metallic
(3) the electric potential decreases surface is zero.
with increasing distance from (c) electric lines of force are
centre. perpendicular to equi-potential
(4) the electric field is perpendicular to surface.
(1) (a) and (b) only
the line joining the charges.
(2) (b) and (c) only
71. An electric dipole is made up of two (3) (a) and (c) only
–6 (4) (a), (b) and (c)
equal and opposite charges of 2 × 10
coulomb at a distance of 3 cm. This is
5
77. The dielectric constant of a metal is:
kept in an electric field of 2 × 10 N/C, (1) ∞ (2) 0
then the maximum torque acting on the (3) 1 (4) none of these
dipole:
–1 –3
(1) 12 × 10 Nm (2) 12 × 10 Nm 78. You are travelling in a car during a
(3) 24 × 10
–3
Nm
–1
(4) 24 × 10 Nm thunder storm, in order to protect
yourself from lightening would you
72. The electric potential in volt due to an prefer to:
–8 (1) Remain in the car.
electric dipole of dipole moment 2 × 10
(2) Take shelter under a tree.
C-m at a distance of 3m on a line making
0
(3) Get out and be flat on the ground
an angle of 60 with the axis of the dipole (4) Touch the nearest electrical pole.
is:
(1) 0 (2) 10 79. Two conductors of the same shape and
(3) 20 (4) 40 size. One of copper and the other of
aluminium (less conducting) are placed
73. If an electric dipole is kept in a non- in an uniform electric field. The charge
uniform electric field, then it will induced in aluminium
experience: (1) will be less than in copper.
(1) only torque (2) will be more than in copper.
(2) no torque (3) will be equal to that in copper.
(3) a resultant force and a torque (4) will not be connected with copper.
(4) only a force
80. The electric field near the conducting
Conductor, surface of a uniform charge density σ
will be -
Sharing of Charge
(1) σ/ ∈0 and parallel to surface.
74. 64 charged drops coalesce to form a (2) 2σ/ ∈0 and parallel to surface.
bigger charged drop. The potential of
bigger drop will be times that of (3) σ/ ∈0 and perpendicular to surface.
smaller drop: (4) 2σ/ ∈0 and perpendicular to surface.
(1) 4 (2) 16
(3) 64 (4) 8 81. A neutral metallic object is placed near
a finite metal plate carrying a positive
75. 27 smaller drop combine to form a charge. The electric force on the object
bigger drop if potential on smaller drop will be:
is V then potential on bigger drop will (1) towards the plate
be- (2) away from the plate
(1) 9V (2) 3V (3) parallel to the plate
(3) 27V (4) 1/3V (4) zero
82. The net charge given to an isolated (3) must be distributed uniformly in the
conducting solid sphere: volume.
(1) must be distributed uniformly on (4) may be distributed uniformly in the
the surface. volume.
(2) may be distributed uniformly on the
surface.

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Ans. 3 4 1 2 2 1 4 2 2 2 3 1 1 2 1 1 4 3 1 4 2 3 3 4 3
Que. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
Ans. 4 3 2 4 4 3 3 3 4 1 3 4 2 4 2 2 4 1 1 4 1 2 3 2 3
Que. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75
Ans. 1 3 1 2 4 4 2 3 1 4 1 3 3 4 2 4 4 1 2 2 2 2 3 2 1
Que. 76 77 78 79 80 81 82
Ans. 2 1 1 3 3 1 1
 Exercise - II
1. An electric dipole is kept on the axis of 5. A charged particle 'q' is shot towards
a uniform charged ring at large distance another charged particle 'Q', which is
from the centre of the ring. The fixed, with a speed 'v'. It approaches 'Q'
direction of the dipole moment is along upto a closest distance r and then
the axis. The dipole moment is p, returns. If q were given a speed of '2v',
the closest distance of approach would
charge of the ring is Q & radius of the
be:
ring is R. The force on the dipole is
q Q
pQ 4pQ v r
(1) (2)
3πε0 3R2 3πε0 3R2 (1) r (2) 2r
(3) r/2 (4) r/4
pQ
(3) (4) Zero
3πε0R2 6. Assume that an electric field E = 30x î
2

exists in space. Then, the potential

2. A thin semi-circular ring of radius r has difference VA – VO, where VO is the
a positive charge q distributed potential at the origin and VA the

uniformly over it. The net field E at the potential at x = 2 m, is-
centre O is: (1) 120 J (2) –120 J
(3) –80 J (4) 80 J
ĵ

7. Two identical charged spheres are

θ suspended by strings of equal lengths.
O
î The strings make an angle of 30º with
each other. When suspended in a liquid
q ˆj q ˆj
–3
of density 0.8 g cm , the angle remains
(1) 2 2
(2) –
4π ε0r 4π ε0r2
2
the same. If density of the material of
–3
the sphere is 1.6 g cm , the dielectric
q ˆj q ˆj
(3) – (4) constant of the liquid is
2π2 ε0r2 2π2 ε0r2 (1) 4 (2) 3
(3) 2 (4) 1
3. A charge Q is placed at each of the
opposite corners of a square. A charge 8. Two points P and Q are maintained at
q is placed at each of the other two the potentials of 10 V and –4V,
respectively. The work done in moving
corners. If the net electrical force on Q
100 electrons from P and Q.
is zero, then Q/q equals: –17
(1) 9.60×10 J
(1) –1 (2) 1 –16
(2) –2.24×10 J
1
(3) – (4) – 2 2 –16
(3) 2.24×10 J
2 –17
(4) –9.60×10 J

4. Four charges equal to –Q each are 9. An electric dipole is placed at an angle

placed at the four corners of a square of 30° to a non – uniform electric field.
and a charge q is at its centre. If the The dipole will experience
system is in equilibrium, the value of q (1) a translational force only in a
is: direction perpendicular to the field
Q Q (2) a torque as well as a translational
(1) – (1 + 2 2) (2) (1 + 2 2) force
4 4
(3) a torque only
Q Q (4) a translational force only in the
(3) – (1 + 2 2) (4) (1 + 2 2)
2 2 direction of the field
10. A particle of charge –q & mass m
(1) 2 qa along + y direction.
moves in a circle of radius r around an
infinitely long line charge of linear charge (2) 2 qa along the line joining points
density +λ, then time period will be. (x = 0, y = 0, z = 0) and (x = a, y = a,
1 z = 0).
Where k = (3) qa along the line joining points (x = 0,
4πε0
y = 0, z = 0) and (x = a, y = a, z = 0).
+λ
(4) 2 qa along + x direction.

r
–q 13. A thin conducting ring of radius R is
given a charge +Q. The electric field at
the centre O of the ring due to the
m charge on the part AKB of the ring is E.
(1) T = 2πr The electric field at the centre due to
2kλq the charge on the part ACDB of the ring
2 4π2m is:
(2) T = A
2kλq K

1 2kλq
(3) T = C
O
B
2πr m
1 m D
(4) T =
2πr 2kλq (1) 3E along KO (2) E along OK
(3) E along KO (4) 3 E along OK
11. Two charges q1 and q2 are placed 30 cm
14. The electric potential at a point (x, y, z)
apart, as shown in the figure. A third 2 3
charge q3 is moved along the arc of a is given by V = – x y – xz + 4

circle of radius 40 cm from C to D. The The electric field E at that point is:

change in the potential energy of the (1) E = î (2xy+ z ) + ˆj x + k̂ 3xz
3 2 2

q3 
(2) E = î 2xy + ˆj (x + y ) + k̂ (3xz – y )
2 2 2
system is k, where k is:
4πε0 
(3) E = î z + ˆj xyz + k̂ z
3 2
q3

(4) E = î (2xy – z ) + ˆj xy + k̂ 3z x
C 3 2 2

40cm
3R
15. The electric field at a distance from
q2
2
q1
D
the centre of a charged conducting
A B
spherical shell of radius R is E. The
(1) 8q2 (2) 8q1
R
(3) 6q2 (4) 6q1 electric field at a distance from the
2
centre of the sphere is:
12. Three point charges +q, –2q and + q are (1) zero (2) E
placed at points (x = 0, y = a, z = 0), (x =
E E
0, y = 0, z = 0) and (x = a, y = 0, z = 0), (3) (4)
respectively. The magnitude and 2 3
direction of the electric dipole moment
vector of this charge assembly are:

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans. 4 3 4 2 4 3 3 3 2 1 1 2 2 1 1
 Exercise – III (Previous Year Question)

1. Charges +q and –q are placed at points qC

A and B respectively which are a
distance 2 L apart, C is the midpoint E D
between A and B. The work done in
moving a charge +Q along the q
q
semicircle CRD is: [AIPMT - 2007] A B
R
3qQ qQ
(1) (2)
8π ∈0 a 4π ∈0 a
A C B D
3qQ
qQ qQ (3) zero (4)
(1) (2) 4π ∈0 a
4πε0L 2πε0L
5. An electric dipole of moment ´p´ is
qQ qQ
(3) (4) – placed in an electric field of intensity
6πε0L 6πε0L ´E´. The dipole acquires a position
such that the axis of the dipole makes
2. A charge Q is enclosed by a Gaussian an angle θ with the direction of the
spherical surface of radius R. If the field. Assuming that the potential
radius is doubled, then the outward energy of the dipole to be zero when
electric flux will: [AIPMT- 2011]
θ = 90°, the torque and the potential
(1) increase four times
energy of the dipole will respectively be:
(2) be reduced to half
[AIPMT_Pre_2012]
(3) remain the same
(4) be doubled (1) p E sin θ, – p E cos θ
(2) p E sin θ, – 2 p E cos θ
3. Four electric charges +q, +q, –q and –q (3) p E sin θ, 2 p Ecos θ
are placed at the corners of a square of (4) p E cos θ, – p Ecos θ
side 2L (see figure). The electric
potential at point A, midway between 6. Four point charges –Q, –q, 2q and 2Q
the two charges +q and +q, is: are placed, one at each corner of the
[AIPMT- 2011] square. The relation between Q and q
+q –q
for which the potential at the centre of
A
the square is zero is: [AIPMT_Pre_2012]
(1) Q = –q (2) Q = – 1/q
+q –q (3) Q = q (4) Q = 1/q
1 2q
(1) (1 + 5) 7. What is the flux through a cube of side
4π ∈0 L 'a' if a point charge of q is at one of its
1 2q  1  corner: [AIPMT_Pre_2012]
(2) 1+  q
4π ∈0 L  5
1 2q  1 
(3) 1– 
4π ∈0 L  5
(4) Zero
4. Three charges, each +q, are placed at a
the corners of an isosceles triangle ABC
2q q
of sides BC and AC, 2a. D and E are the (1) (2)
mid points of BC and CA. The work ε0 8ε0
done in taking a charge Q from D to E q q
is: [AIPMT Mains 2011] (3) (4) 6a2
ε0 2ε0
8. Two metallic spheres of radii 1 cm and 3 12. In a region, the potential is represented
–2
cm are given charges of –1 × 10 C and 5 by V (x, y, z) = 6x – 8xy –8y + 6yz,
–2 where V is in volts and x, y, z are in
× 10 C, respectively. If these are
metres. The electric force experienced
connected by a conducting wire, the
by a charge of 2 coulomb situated at
final charge on the bigger sphere is:
point (1, 1, 1) is: [AIPMT-2014]
[AIPMT 2012 (Mains)]
–2
(1) 2 × 10 C
–2
(2) 3 × 10 C (1) 6 5N (2) 30 N
(4) 4 35N
–2 –2
(3) 4 × 10 C (4) 1 × 10 C (3) 24 N
13. The electric field in a certain region is
9. Two pith balls carrying equal charges acting radially outward and is given by
are suspended from a common point E = Ar. A charge contained in a sphere
by strings of equal length, the of radius 'a' centered at the origin of
equilibrium separation between them is the field, will be given by:
r. Now the strings are rigidly clamped [AIPMT-2015]
2 3
at half the height. The equilibrium (1) A ε0 a (2) 4 πε0 Aa
3 2
separation between the balls now (3) ε0 Aa (4) 4 πε0 Aa
become: [NEET_2013]
14. If potential (in volts) in a region is
expressed as V (x, y, z) = 6xy – y + 2yz,
y
y/2
the electric field (in N/C) at point (1, 1, 0)
r r' is: [AIPMT-2015]
 r 
(1)  
 2r 
(2)  
( ˆ
(1) – 6iˆ + 9ˆj + k ) (
(2) – 3iˆ + 5ˆj + 3kˆ )
(3) – ( 6iˆ + 5ˆj + 2kˆ ) (4) – ( 2iˆ + 3ˆj + k
ˆ)
3
 2  3
2
 2r   r 
(3)   (4)  
3 15. Two identical charged spheres
 2
suspended from a common point by
10. A, B and C are three points in a uniform two massless strings of lengths , are
electric field. The electric potential is: initially at a distance (d<<) apart
[NEET_2013] because to their mutual repulsion. The
B A  charges begin to leak from both the
E spheres at a constant rate. As a result,
C
the spheres approach each other with a
(1) maximum at B velocity v. Then v varies as a function
(2) maximum at C of the distance x between the spheres,
(3) same at all the three points A, B and C as: [NEET - 2016]
1
(4) maximum at A
(1) v ∝ x 2 (2) v ∝ x
1
–
11. A conducting sphere of radius R is given (3) v ∝ x 2 (4) v ∝ x–1
a charge Q. The electric potential and
the electric field at the centre of the 16. Suppose the charge of a proton and an
sphere respectively are: [AIPMT-2014] electron differ slightly. One of them is
Q –e, the other is (e +∆e). If the net of
(1) Zero and electrostatic force and gravitational
4π ∈0 R2
force between two hydrogen atoms
Q placed at a distance d (much greater
(2) and zero
4π ∈0 R than atomic size) apart is zero, then ∆e is
of the order of [Given mass of hydrogen
Q Q –27
(3) and mh = 1.67 × 10 kg] [NEET - 2017]
4π ∈0 R 4π ∈0 R2 –20 –23
(1) 10 C (2) 10 C
(4) Both are zero –37 –47
(3) 10 C (4) 10 C
17. The diagrams below show regions of 20. Two point charges A and B, having
equipotential surfaces. charges +Q and –Q respectively, are
20V 40V 20V 40V 10V 30V
40V placed at certain distance apart and
20V
force acting between them is F. If 25%
A B A B A B A B charge of A is transferred to B, then
10V force between the charges becomes:
30V
10V 30V 10V 30V 20V 40V [NEET-2019]
(a) (b) (c) (d)
16F 4F
(1) (2)
A positive charge is moved from A to B 9 3
in each diagram, then for q moving
9F
from A to B: [NEET - 2017] (3) F (4)
16
(1) Maximum work is required to move
q in figure (c). 21. A hollow metal sphere of radius R is
(2) In all the four cases the work done uniformly charged. The electric field
is the same. due to the sphere at a distance r from
(3) Minimum work is required to move the centre: [NEET-2019]
q in figure (a) (1) zero as r increases for r < R,
(4) Maximum work is required to move increases as r increases for r > R
q in figure (b). (2) decreases as r increases for r < R
and for r > R
18. An electron falls from rest through a (3) increases as r increases for r < R
vertical distance h in a uniform and and for r > R
vertically upward directed electric field (4) zero as r increases for r < R,
E. The direction of electrical field is decreases as r increases for r > R
now reversed, keeping its magnitude
the same. A proton is allowed to fall 22. Two parallel infinite line charges with
from rest in through the same vertical linear charge density +λ C/m and –λ
distance h. The time fall of the C/m are placed at a distance of 2R in
electron, in comparison to the time fall free space. What is the electric field
of the proton is [NEET-2018] mid-way between the two lines charges
(1) smaller [NEET-2019]
(2) 5 times greater λ λ
(1) N/C (2) N/C
(3) 10 times greater π ∈0 R 2π ∈0 R
(4) equal
2λ
(3) zero (4) N/C
π ∈0 R
19. A toy car with charge q moves on a
frictionless horizontal plane surface 23. Two metal spheres, one of radius R and
under the influence of a uniform the other of radius 2R respectively have
electric field E Due to the force qE, its the same surface charge density σ.
velocity increases from 0 to 6 m/sec in They are brought in contact and
one second duration. At that instant separated. What will be the new
the direction of the field is reversed. surface charge densities on them
The car continues to move for two [NEET-2019 Odisha]
more seconds under the influence of 5 5
(1) σ1 = σ, σ2 = σ
this field. The average velocity and the 6 2
average speed of the toy car between 0 5 5
to 3 seconds are respectively (2) σ1 = σ, σ2 = σ
2 6
[NEET-2018]
(1) 2 m/sec, 4 m/sec 5 5
(3) σ1 = σ, σ2 = σ
(2) 1 m/sec, 3 m/sec 2 3
(3) 1 m/sec, 3.5 m/sec 5 5
(4) σ1 = σ, σ2 = σ
(4) 1.5 m/sec, 3 m/sec 3 6
24. A sphere encloses an electric dipole 29. The acceleration of an electron due to
with charge ± 3 × 10–6 C. What is the the mutual attraction between the
total electric flux across the sphere: electron and a proton when they are 1.6
[NEET-2019 Odisha] Å a part is, (me ≈ 9 × 10–31 kg, e = 1.6 ×
(1) – 3 × 10–6 Nm2/C 1
(2) zero 10–19 C) (Take = 9 × 109 Nm2 C–2 )
4πε0
(3) 3 × 10–6 Nm2/C
[NEET (Covid) - 2020]
(4) 6 × 10–6 Nm2/C
(1) 1024 m/s2 (2) 1023 m/s2
(3) 1022 m/s2 (4) 1025 m/s2
25. A spherical conductor of radius 10 cm
has a charge of 3.2 × 10–7 C distributed
30. The variation of electrostatic potential
uniformly. what is the magnitude of
with radial distance r from the centre
electric field at a point 15 cm from the
of a positively charged metallic thin
centre of the sphere? [NEET - 2020]
shell of radius R is given by the graph
 1 
 = 9 × 109 Nm2 / C2  [NEET (Covid) - 2020]
 4πε 0 
(1) 1.28 × 106 N/C (2) 1.28 × 107 N/C
V
(3) 1.28 × 104 N/C (4) 1.28 × 105 N/C
(1)
26. In a certain region of space with r
volume 0.2 m3, the electric potential is R
found to be 5 V throughout. The
magnitude of electric field in this region
is: [NEET - 2020] V

(1) 1 N/C (2) 5 N/C

(2)
(3) Zero (4) 0.5 N/C
r
R
27. A short electric dipole has a dipole
moment of 16 × 10–9 C m. the electric
potential due to the dipole at a point at V
a distance of 0.6 m from the centre of
the dipole, situated on a line making an (3)
angle of 60° with the dipole axis is: r
R
 1 
 = 9 × 109 Nm2 / C2  [NEET - 2020]
 4πε0 
(1) 400 V (2) Zero V
(3) 50 V (4) 200 V
(4)
28. The electric field at a point on the r
equatorial plane at a distance r from R
the centre of a dipole having dipole 31. Two charged spherical conductors of

moment P is given by (r >> separation radius R1 andR2 are connected by a
wire. Then the ratio of surface charge
of two charges forming the dipole, ∈0
permittivity of free space) densities of the spheres (σ1/σ2) is:
[NEET (Covid) - 2020] [NEET - 2021]
  R R
 P  2P (A) 1 (B) 2
(1) E = (2) E = R2 R1
4πε0r3 4πε0r3
   R1  R21
 P  P (C)   (D)
(3) E = – (4) E = –
4πε0r2 4π ε0 r3  R2  R22
32. A dipole is placed in an electric field as −q2
shown. In which direction will it move? (1) Zero (2)
4πε0d
[NEET - 2021]
−q2  1  q2  1 
(3) 6 −  (4) 6 − 
4πε0d  2 4πε0d  2

E 35. The ratio of coulomb's electrostatic
+ –
force to the gravitational force between
an electron and a proton separated by
some distance is 2.4 × 1039. The ratio of
(A) towards the left as its potential 1
energy will increase. the proportionality constant, K =
4πε0
(B) towards the right as its potential
energy will decrease. to the Gravitational constant G is nearly
(C) towards the left as its potential (Given that the charge of the proton
energy will decrease. and electron each = 1.6 × 10–19 C, the
(D) towards the right as its potential mass of the electron = 9.11 × 10–31 kg,
energy will increase. the mass of the proton = 1.67 × 10–27 kg :)
[NEET - 2022]
33. Twenty seven drops of same size are (1) 1020 (2) 1030
charged at 220 V each. They combine (3) 1040 (4) 10
to form a bigger drop. Calculate the
potential of the bigger drop.
[NEET - 2021]
(1) 660 V (2) 1320 V
(3) 1520 V (4) 1980 V

34. Six charges +q, –q, +q, –q, +q and –q

are fixed at the corners of a hexagon of
side d as shown in the figure. The work
done in bringing a charge q0 to the
centre of the hexagon from infinity is :
[NEET - 2022]

ANSWER KEY
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
Ans. 4 3 3 3 1 1 2 2 1 1 2 4 2 3 3 3 2 1 2 4 4 1 4 2 4
Que. 26 27 28 29 30 31 32 33 34 35
Ans. 3 4 4 3 2 2 2 4 1 1